MHT CET 2025 Mathematics Question Paper with Answer and Solution

(C) Let $I = \int \frac{dx}{x(x^3+1)}$.
Multiply the numerator and denominator by $x^2$:
$I = \int \frac{x^2 dx}{x^3(x^3+1)}$.
Let $x^3 = t$,then $3x^2 dx = dt$,or $x^2 dx = \frac{dt}{3}$.
Substituting these into the integral:
$I = \frac{1}{3} \int \frac{dt}{t(t+1)}$.
Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.
$I = \frac{1}{3} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt$.
$I = \frac{1}{3} (\log|t| - \log|t+1|) + c$.
$I = \frac{1}{3} \log \left|\frac{t}{t+1}\right| + c$.
Substituting $t = x^3$ back:
$I = \frac{1}{3} \log \left|\frac{x^3}{x^3+1}\right| + c = \log \left|\left(\frac{x^3}{x^3+1}\right)^{1/3}\right| + c = \log \left(\sqrt[3]{\frac{x^3}{x^3+1}}\right) + c$.

(C) Let $I = \int \frac{dx}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}$.
Rewrite the integrand as: $I = \int \frac{dx}{(x+a)^{\frac{9}{7}} \cdot (x+a)^{\frac{5}{7}} \cdot \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}} = \int \frac{dx}{(x+a)^2 \left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}}$.
Let $t = \frac{x-b}{x+a}$. Then $dt = \frac{(x+a)(1) - (x-b)(1)}{(x+a)^2} dx = \frac{a+b}{(x+a)^2} dx$.
Thus,$\frac{dx}{(x+a)^2} = \frac{dt}{a+b}$.
Substituting these into the integral,we get: $I = \int \frac{1}{t^{\frac{5}{7}}} \cdot \frac{dt}{a+b} = \frac{1}{a+b} \int t^{-\frac{5}{7}} dt$.
Integrating,we get: $I = \frac{1}{a+b} \cdot \frac{t^{-\frac{5}{7}+1}}{-\frac{5}{7}+1} + c = \frac{1}{a+b} \cdot \frac{t^{\frac{2}{7}}}{\frac{2}{7}} + c = \frac{7}{2(a+b)} \left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + c$.

(B) Let $I = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \,dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\tan x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \,dx = \int \frac{\sqrt{\tan x} \cdot \sec^2 x}{\tan x} \,dx$.
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} \,dx$.
Let $u = \tan x$, then $du = \sec^2 x \,dx$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{u}} \,du = \int u^{-1/2} \,du$.
Using the power rule $\int u^n \,du = \frac{u^{n+1}}{n+1} + c$:
$I = \frac{u^{1/2}}{1/2} + c = 2 \sqrt{u} + c$.
Substituting back $u = \tan x$:
$I = 2 \sqrt{\tan x} + c$.

(A) To solve the integral $I = \int \frac{dx}{x^{1/2} + x^{1/3}}$,we use the substitution $x = t^6$,so $dx = 6t^5 dt$.
Substituting these into the integral,we get:
$I = \int \frac{6t^5 dt}{t^3 + t^2} = \int \frac{6t^5}{t^2(t+1)} dt = \int \frac{6t^3}{t+1} dt$.
Using polynomial division,$\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$.
Thus,$I = 6 \int (t^2 - t + 1 - \frac{1}{t+1}) dt = 6 [\frac{t^3}{3} - \frac{t^2}{2} + t - \log|t+1|] + k$.
$I = 2t^3 - 3t^2 + 6t - 6 \log|t+1| + k$.
Substituting $t = x^{1/6}$ back,we get:
$I = 2(x^{1/6})^3 - 3(x^{1/6})^2 + 6(x^{1/6}) - 6 \log(x^{1/6} + 1) + k$.
$I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log(x^{1/6} + 1) + k$.
Comparing this with the given form $A x^{1/2} + B x^{1/3} + C x^{1/6} + D \log(x^{1/6} + 1) + k$,we find $A = 2, B = -3, C = 6, D = -6$.

(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$.
Substitute $u = e^x$,then $du = e^x dx$.
The integral becomes $I = \int \frac{du}{\sqrt{u^2+4u+13}}$.
Complete the square for the quadratic expression: $u^2+4u+13 = (u+2)^2 + 9 = (u+2)^2 + 3^2$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + c$,we get:
$I = \log |(u+2) + \sqrt{(u+2)^2 + 3^2}| + c$.
Substituting $u = e^x$ back,we get:
$I = \log |e^x + 2 + \sqrt{e^{2x}+4e^x+13}| + c$.
Comparing this with the given expression $\log |e^{ax}+2+\sqrt{e^{2x}+4e^x+13}| + c$,we find that $a = 1$.

(C) We have the integral $I = \int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$.
Since $7x = 9x - 2x$, we can write $\sin 7x = \sin(9x - 2x)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we get:
$I = \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x} \,d x$.
Splitting the fraction:
$I = \int \left( \frac{\sin 9x \cos 2x}{\cos 9x \cos 2x} - \frac{\cos 9x \sin 2x}{\cos 9x \cos 2x} \right) \,d x$.
$I = \int (\tan 9x - \tan 2x) \,d x$.
Integrating term by term:
$I = \int \tan 9x \,d x - \int \tan 2x \,d x$.
Using the formula $\int \tan(ax) \,d x = \frac{1}{a} \ln |\sec(ax)| + c$:
$I = \frac{1}{9} \ln |\sec 9x| - \frac{1}{2} \ln |\sec 2x| + c$.
Thus, the correct option is $C$.

(C) Let $I = \int \frac{dx}{\cos x(1+\cos x)}$.
We can write the integrand as $\frac{1}{\cos x(1+\cos x)} = \frac{1+\cos x - \cos x}{\cos x(1+\cos x)} = \frac{1}{\cos x} - \frac{1}{1+\cos x}$.
Now,$I = \int \sec x \, dx - \int \frac{dx}{1+\cos x}$.
We know that $\int \sec x \, dx = \log |\sec x + \tan x| + c_1$.
For the second part,use the identity $1+\cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.
So,$\int \frac{dx}{1+\cos x} = \int \frac{dx}{2 \cos^2 \left(\frac{x}{2}\right)} = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) dx$.
Integrating this,we get $\frac{1}{2} \cdot \frac{\tan(x/2)}{1/2} = \tan \left(\frac{x}{2}\right) + c_2$.
Combining these,$I = \log |\sec x + \tan x| - \tan \left(\frac{x}{2}\right) + c$.

(C) We want to evaluate the integral $I = \int e^x \frac{x-1}{(x+1)^3} \, dx$.
Rewrite the numerator as $(x+1) - 2$:
$I = \int e^x \frac{(x+1) - 2}{(x+1)^3} \, dx$
$I = \int e^x \left( \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right) \, dx$
$I = \int e^x \left( \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right) \, dx$
Recall the standard integral form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + c$.
Let $f(x) = \frac{1}{(x+1)^2} = (x+1)^{-2}$.
Then $f'(x) = -2(x+1)^{-3} = -\frac{2}{(x+1)^3}$.
Since the integrand is in the form $e^x [f(x) + f'(x)]$,the integral is $e^x f(x) + c$.
Therefore,$I = \frac{e^x}{(x+1)^2} + c$.

(A) Let $I = \int \log (2+x)^{2+x} \, dx$.
Using the property $\log(a^b) = b \log a$,we get:
$I = \int (2+x) \log(2+x) \, dx$.
Let $u = 2+x$,then $du = dx$.
The integral becomes $I = \int u \log u \, du$.
Using integration by parts $\int f(u)g'(u) \, du = f(u)g(u) - \int f'(u)g(u) \, du$,let $f(u) = \log u$ and $g'(u) = u$.
Then $f'(u) = \frac{1}{u}$ and $g(u) = \frac{u^2}{2}$.
$I = \frac{u^2}{2} \log u - \int \frac{1}{u} \cdot \frac{u^2}{2} \, du$
$I = \frac{u^2}{2} \log u - \frac{1}{2} \int u \, du$
$I = \frac{u^2}{2} \log u - \frac{1}{2} \cdot \frac{u^2}{2} + c$
$I = \frac{u^2}{2} \left( \log u - \frac{1}{2} \right) + c$
Since $\frac{1}{2} = \log \sqrt{e}$,we have:
$I = \frac{u^2}{2} (\log u - \log \sqrt{e}) + c = \frac{u^2}{2} \log \left( \frac{u}{\sqrt{e}} \right) + c$.
Substituting $u = 2+x$ back:
$I = \frac{(2+x)^2}{2} \log \left( \frac{2+x}{\sqrt{e}} \right) + c$.

(D) Let $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} \, dx$.
We can rewrite the integrand as:
$I = \int e^{2x} \left( \frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} \right) \, dx$
$I = \int e^{2x} (\cot 2x - \csc^2 2x) \, dx$.
Let $f(x) = \cot 2x$. Then $f'(x) = -\csc^2 2x \cdot 2 = -2 \csc^2 2x$.
This does not fit the form $\int e^{ax} (f(x) + \frac{f'(x)}{a}) \, dx$.
Let us rewrite the integral as $I = \int e^{2x} \cot 2x \, dx - \int e^{2x} \csc^2 2x \, dx$.
Using integration by parts on $\int e^{2x} \cot 2x \, dx$:
Let $u = \cot 2x$,$dv = e^{2x} \, dx$. Then $du = -2 \csc^2 2x \, dx$ and $v = \frac{1}{2} e^{2x}$.
$I = (\cot 2x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (-2 \csc^2 2x) \, dx) - \int e^{2x} \csc^2 2x \, dx$.
$I = \frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x \, dx - \int e^{2x} \csc^2 2x \, dx + c$.
$I = \frac{1}{2} e^{2x} \cot 2x + c$.

(D) We use the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
First,rewrite the integrand: $\frac{x+5}{(x+6)^2} = \frac{(x+6) - 1}{(x+6)^2} = \frac{1}{x+6} - \frac{1}{(x+6)^2}$.
Let $f(x) = \frac{1}{x+6}$.
Then $f'(x) = -\frac{1}{(x+6)^2}$.
Substituting these into the integral: $\int e^x \left( \frac{1}{x+6} - \frac{1}{(x+6)^2} \right) dx = \int e^x [f(x) + f'(x)] dx$.
Applying the formula,we get $e^x f(x) + c = \frac{e^x}{x+6} + c$.

(A) To evaluate $\int x^2 \cos x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x^2$ and $dv = \cos x \, dx$.
Then $du = 2x \, dx$ and $v = \sin x$.
Applying the formula:
$\int x^2 \cos x \, dx = x^2 \sin x - \int (\sin x)(2x) \, dx = x^2 \sin x - 2 \int x \sin x \, dx$.
Now,apply integration by parts again for $\int x \sin x \, dx$:
Let $u = x$ and $dv = \sin x \, dx$.
Then $du = dx$ and $v = -\cos x$.
$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x$.
Substituting this back into the original expression:
$\int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) + c = x^2 \sin x + 2x \cos x - 2 \sin x + c$.

(A) Let $I = \int e^x \cos x \, dx$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = \cos x$ and $dv = e^x \, dx$.
Then $du = -\sin x \, dx$ and $v = e^x$.
$I = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$.
Now,evaluate $\int e^x \sin x \, dx$ using integration by parts again.
Let $u = \sin x$ and $dv = e^x \, dx$.
Then $du = \cos x \, dx$ and $v = e^x$.
$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$.
Substituting this back into the equation for $I$:
$I = e^x \cos x + e^x \sin x - I$.
$2I = e^x (\cos x + \sin x)$.
$I = \frac{e^x (\cos x + \sin x)}{2} + c$.

(B) We know that $\tan^{-1} \left( \frac{2x}{1-x^2} \right) = 2 \tan^{-1} x$ for $|x| < 1$.
Substituting this into the integral,we get $I = \int 2 \tan^{-1} x \, dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = 2 \, dx$.
Then $du = \frac{1}{1+x^2} \, dx$ and $v = 2x$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I = 2x \tan^{-1} x - \int \frac{2x}{1+x^2} \, dx$.
Let $t = 1+x^2$,then $dt = 2x \, dx$.
So,$\int \frac{2x}{1+x^2} \, dx = \int \frac{1}{t} \, dt = \log|t| + c = \log(1+x^2) + c$.
Thus,$I = 2x \tan^{-1} x - \log(1+x^2) + c$.
Rearranging the terms,we get $I - 2x \tan^{-1} x = -\log(1+x^2) + c$.

(C) To evaluate the integral $\int \frac{x \, dx}{(x-1)(x-2)}$,we use partial fractions.
Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
Multiplying both sides by $(x-1)(x-2)$,we get $x = A(x-2) + B(x-1)$.
Setting $x = 1$,we get $1 = A(1-2) \implies A = -1$.
Setting $x = 2$,we get $2 = B(2-1) \implies B = 2$.
Thus,$\int \frac{x \, dx}{(x-1)(x-2)} = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$.
Integrating term by term,we get $-\log |x-1| + 2 \log |x-2| + c$.
Using the property of logarithms,$n \log m = \log m^n$,we get $\log |x-2|^2 - \log |x-1| + c = \log \left| \frac{(x-2)^2}{x-1} \right| + c$.

(B) To solve the integral $\int \frac{2x+3}{(x-1)(x^2+1)} dx$,we use partial fractions:
$\frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1}$
$2x+3 = B(x^2+1) + (Cx+D)(x-1)$
Setting $x=1$,we get $2(1)+3 = B(1^2+1) \implies 5 = 2B \implies B = \frac{5}{2}$.
Comparing coefficients of $x^2$: $0 = B+C \implies C = -B = -\frac{5}{2}$.
Comparing constants: $3 = B-D \implies D = B-3 = \frac{5}{2} - 3 = -\frac{1}{2}$.
Thus,the integral becomes $\int \left( \frac{5/2}{x-1} + \frac{-5/2x - 1/2}{x^2+1} \right) dx$.
$= \frac{5}{2} \int \frac{1}{x-1} dx - \frac{5}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
$= \frac{5}{2} \log_e |x-1| - \frac{5}{4} \log_e (x^2+1) - \frac{1}{2} \tan^{-1} x + A$.
$= \log_e |x-1|^{\frac{5}{2}} + \log_e (x^2+1)^{-\frac{5}{4}} - \frac{1}{2} \tan^{-1} x + A$.
$= \log_e \{(x-1)^{\frac{5}{2}}(x^2+1)^{-\frac{5}{4}}\} - \frac{1}{2} \tan^{-1} x + A$.
Comparing this with the given expression,we find $a = -\frac{5}{4}$.

(C) Let $I = \int \frac{3 \sin x \cos x}{4 \sin x + 7} \, dx$.
Substitute $u = \sin x$,then $du = \cos x \, dx$.
The integral becomes $I = \int \frac{3u}{4u+7} \, du$.
Perform polynomial division or rewrite the numerator: $\frac{3u}{4u+7} = \frac{3}{4} \left( \frac{4u}{4u+7} \right) = \frac{3}{4} \left( \frac{4u+7-7}{4u+7} \right) = \frac{3}{4} \left( 1 - \frac{7}{4u+7} \right) = \frac{3}{4} - \frac{21}{4(4u+7)}$.
Integrating with respect to $u$: $I = \int \left( \frac{3}{4} - \frac{21}{4(4u+7)} \right) \, du = \frac{3}{4}u - \frac{21}{4} \cdot \frac{1}{4} \log |4u+7| + c$.
Substituting back $u = \sin x$: $I = \frac{3}{4} \sin x - \frac{21}{16} \log |4 \sin x + 7| + c$.
Comparing this with $A \sin x - B \log |4 \sin x + 7| + c$,we get $A = \frac{3}{4}$ and $B = \frac{21}{16}$.
Thus,$A+B = \frac{3}{4} + \frac{21}{16} = \frac{12+21}{16} = \frac{33}{16}$.

(B) Let $I = \int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx$.
Using partial fractions,let $x^2 = t$. Then $\frac{2t+3}{(t-1)(t-4)} = \frac{A}{t-1} + \frac{B}{t-4}$.
$2t+3 = A(t-4) + B(t-1)$.
For $t=1$,$5 = A(-3) \implies A = -\frac{5}{3}$.
For $t=4$,$11 = B(3) \implies B = \frac{11}{3}$.
So,$\frac{2x^2+3}{(x^2-1)(x^2-4)} = \frac{-5/3}{x^2-1} + \frac{11/3}{x^2-4}$.
Integrating,$I = -\frac{5}{3} \int \frac{dx}{x^2-1} + \frac{11}{3} \int \frac{dx}{x^2-4}$.
$I = -\frac{5}{3} \cdot \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| + \frac{11}{3} \cdot \frac{1}{4} \log \left| \frac{x-2}{x+2} \right| + c$.
$I = \frac{5}{6} \log \left| \frac{x+1}{x-1} \right| + \frac{11}{12} \log \left| \frac{x-2}{x+2} \right| + c$.
Comparing with the given form,$a = \frac{11}{12}$ and $b = \frac{5}{6} = \frac{10}{12}$.
Thus,$a+b = \frac{11}{12} + \frac{10}{12} = \frac{21}{12}$.

(A) Let $I = \int \frac{dx}{x^4+5x^2+4}$.
Factor the denominator: $x^4+5x^2+4 = (x^2+1)(x^2+4)$.
Using partial fractions: $\frac{1}{(x^2+1)(x^2+4)} = \frac{A'}{x^2+1} + \frac{B'}{x^2+4}$.
$1 = A'(x^2+4) + B'(x^2+1)$.
Setting $x^2 = -1$,we get $1 = A'(3) \implies A' = \frac{1}{3}$.
Setting $x^2 = -4$,we get $1 = B'(-3) \implies B' = -\frac{1}{3}$.
So,$I = \int \left( \frac{1/3}{x^2+1} - \frac{1/3}{x^2+4} \right) dx$.
$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.
$I = \frac{1}{3} \tan^{-1} x - \frac{1}{3} \cdot \frac{1}{2} \tan^{-1} \frac{x}{2} + c$.
$I = \frac{1}{3} \tan^{-1} x - \frac{1}{6} \tan^{-1} \frac{x}{2} + c$.
Comparing with $A \tan^{-1} x + B \tan^{-1} \frac{x}{2} + c$,we get $A = \frac{1}{3}$ and $B = -\frac{1}{6}$.

(A) We have the integral $I = \int \frac{x^4+1}{x(x^2+1)^2} dx$.
Multiply the numerator and denominator by $x$:
$I = \int \frac{x(x^4+1)}{x^2(x^2+1)^2} dx$.
Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{dt}{2}$.
Substituting this into the integral:
$I = \frac{1}{2} \int \frac{t^2+1}{t(t+1)^2} dt$.
Using partial fractions: $\frac{t^2+1}{t(t+1)^2} = \frac{A_1}{t} + \frac{B_1}{t+1} + \frac{C_1}{(t+1)^2}$.
$t^2+1 = A_1(t+1)^2 + B_1 t(t+1) + C_1 t$.
For $t=0$,$1 = A_1(1)^2 \implies A_1 = 1$.
For $t=-1$,$2 = C_1(-1) \implies C_1 = -2$.
Comparing coefficients of $t^2$: $1 = A_1 + B_1 \implies 1 = 1 + B_1 \implies B_1 = 0$.
So,$I = \frac{1}{2} \int \left( \frac{1}{t} - \frac{2}{(t+1)^2} \right) dt = \frac{1}{2} \left( \log |t| + \frac{2}{t+1} \right) + c$.
$I = \frac{1}{2} \log |x^2| + \frac{1}{x^2+1} + c = \log |x| + \frac{1}{x^2+1} + c$.
Comparing with $A \log |x| + \frac{B}{1+x^2} + c$,we get $A = 1$ and $B = 1$.
Thus,$A-B = 1-1 = 0$.

(A) We have $I = \int \tan^4 x dx = \int \tan^2 x (\sec^2 x - 1) dx$.
$I = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx$.
$I = \int \tan^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx$.
$I = \frac{\tan^3 x}{3} - \tan x + x + k$.
Comparing this with $a \tan^3 x + b \tan x + c x + k$,we get $a = \frac{1}{3}$,$b = -1$,and $c = 1$.
Therefore,$a - b + c = \frac{1}{3} - (-1) + 1 = \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{7}{3}$.

(A) We use the properties of inverse trigonometric functions:
$1$. $\sin ^{-1}(-x) = -\sin ^{-1}(x)$
$2$. $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$
$3$. $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$
$4$. $\tan ^{-1}(x)$ is standard.
Evaluating each term:
$\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$
$\cos ^{-1}\left(-\frac{1}{2}\right) = \pi - \cos ^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\tan ^{-1}(\sqrt{3}) = \frac{\pi}{3}$
Substituting these values into the expression:
$-\frac{\pi}{4} + \frac{2\pi}{3} - \frac{2\pi}{3} + \frac{\pi}{3} = -\frac{\pi}{4} + \frac{\pi}{3} = \frac{-\pi + 4\pi}{12} = \frac{\pi}{12}$
Thus,the correct option is $A$.

(A) We are given the expression $\cot ^{-1}\left(2 \cos \left(2 \operatorname{cosec}^{-1}(\sqrt{2})\right)\right)$.
First,evaluate $\operatorname{cosec}^{-1}(\sqrt{2})$.
Since $\operatorname{cosec}(\frac{\pi}{4}) = \sqrt{2}$,we have $\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$.
Now,substitute this into the expression:
$2 \cos \left(2 \times \frac{\pi}{4}\right) = 2 \cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the expression becomes $2 \times 0 = 0$.
Finally,we need to find $\cot ^{-1}(0)$.
Since $\cot \left(\frac{\pi}{2}\right) = 0$,we have $\cot ^{-1}(0) = \frac{\pi}{2}$.
Thus,the correct option is $A$.

(A) Let $y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$.
Substitute $x = \tan \theta$,so $\theta = \tan^{-1} x$.
Then $y = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$.
$y = \tan^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}(\tan(\theta/2))$.
$y = \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.

(C) We are given the equation $4 \sin ^{-1} x + \cos ^{-1} x = \pi$.
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the given equation:
$4 \sin ^{-1} x + (\frac{\pi}{2} - \sin ^{-1} x) = \pi$
$3 \sin ^{-1} x + \frac{\pi}{2} = \pi$
$3 \sin ^{-1} x = \frac{\pi}{2}$
$\sin ^{-1} x = \frac{\pi}{6}$
Taking the sine of both sides:
$x = \sin(\frac{\pi}{6}) = \frac{1}{2}$.
Thus,the correct option is $C$.

(C) Let $x = \cos \theta$. Then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ in the expression,we get $\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \tan(\theta/2)$.
So,the expression becomes $y = \sin^2 \left( \cot^{-1} (\tan(\theta/2)) \right)$.
Since $\tan(\theta/2) = \cot(\frac{\pi}{2} - \frac{\theta}{2})$,we have $y = \sin^2 \left( \cot^{-1} \left( \cot \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) \right)$.
This simplifies to $y = \sin^2 \left( \frac{\pi}{2} - \frac{\theta}{2} \right) = \cos^2(\theta/2)$.
Using the identity $\cos^2(\theta/2) = \frac{1+\cos \theta}{2}$,we get $y = \frac{1+x}{2} = \frac{1}{2} + \frac{x}{2}$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2}$.

(B) Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
Substituting this into the expression:
$\tan^{-1}\left(\frac{1-\tan^2 \theta}{2 \tan \theta}\right) = \tan^{-1}\left(\frac{1}{\tan 2\theta}\right) = \tan^{-1}(\cot 2\theta) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta$.
Next,$\cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$.
Adding these two parts: $(\frac{\pi}{2} - 2\theta) + 2\theta = \frac{\pi}{2}$.
Finally,$\sin\left(\frac{\pi}{2}\right) = 1$.

(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{1}{y}\right)$.
This means $\frac{1}{x}=\frac{1}{y}$,so $x=y$.
Substitute $x=y$ into the first equation: $\sin ^{-1} x+\sin ^{-1} x=\frac{\pi}{3}$.
$2 \sin ^{-1} x=\frac{\pi}{3} \implies \sin ^{-1} x=\frac{\pi}{6}$.
Therefore,$x=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.
Since $x=y$,we have $y=\frac{1}{2}$.
Now,calculate $2 x^2+y^2-x y$:
$2 \left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = 2 \left(\frac{1}{4}\right)+\frac{1}{4}-\frac{1}{4} = \frac{1}{2}+0 = \frac{1}{2}$.

(C) Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
For $x = \sqrt{3}$,$\theta = \frac{\pi}{3}$.
The expression becomes $y = \sin^{-1}(\sin 2\theta) + \sec^{-1}(\sec 2\theta)$.
Since $x = \sqrt{3} > 1$,we must be careful with the range.
For $x > 1$,$\sin^{-1}(\frac{2x}{1+x^2}) = \pi - 2\tan^{-1} x$ and $\sec^{-1}(\frac{1+x^2}{1-x^2}) = \pi + 2\tan^{-1} x$ is not correct here.
Let's use the derivative directly:
$\frac{d}{dx}(\sin^{-1}(\frac{2x}{1+x^2})) = \frac{2}{1+x^2}$ for $|x| < 1$,and $-\frac{2}{1+x^2}$ for $|x| > 1$.
$\frac{d}{dx}(\sec^{-1}(\frac{1+x^2}{1-x^2})) = \frac{d}{dx}(\cos^{-1}(\frac{1-x^2}{1+x^2})) = -\frac{2}{1+x^2}$ for $x > 1$.
Thus,for $x > 1$,$\frac{dy}{dx} = -\frac{2}{1+x^2} - \frac{2}{1+x^2} = -\frac{4}{1+x^2}$.
At $x = \sqrt{3}$,$\frac{dy}{dx} = -\frac{4}{1+3} = -\frac{4}{4} = -1$.
However,checking the standard identity $\sec^{-1}(\frac{1+x^2}{1-x^2}) = \cos^{-1}(\frac{1-x^2}{1+x^2})$,for $x > 1$,this is $2\tan^{-1} x - \pi$.
So $y = (\pi - 2\tan^{-1} x) + (2\tan^{-1} x - \pi) = 0$.
Therefore,$\frac{dy}{dx} = 0$.

(B) Let $u = \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$. Put $x = \tan \theta$,then $u = \tan ^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2} \tan ^{-1} x$.
Thus,$\frac{du}{dx} = \frac{1}{2(1+x^2)}$.
Let $v = \tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$. Put $x = \sin \phi$,then $v = \tan ^{-1}\left(\frac{2 \sin \phi \cos \phi}{1-2 \sin^2 \phi}\right) = \tan ^{-1}(\tan 2\phi) = 2\phi = 2 \sin ^{-1} x$.
Thus,$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{2(1+x^2)} \times \frac{\sqrt{1-x^2}}{2} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$.
At $x=0$,$\frac{du}{dv} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

(C) Let $\theta_1 = \sec ^{-1} 4$,then $\sec \theta_1 = 4$.
Using the identity $\tan ^2 \theta = \sec ^2 \theta - 1$,we get $\tan ^2(\sec ^{-1} 4) = \sec ^2(\sec ^{-1} 4) - 1 = 4^2 - 1 = 16 - 1 = 15$.
Let $\theta_2 = \operatorname{cosec}^{-1} 3$,then $\operatorname{cosec} \theta_2 = 3$.
Using the identity $\cot ^2 \theta = \operatorname{cosec}^2 \theta - 1$,we get $\cot ^2(\operatorname{cosec}^{-1} 3) = \operatorname{cosec}^2(\operatorname{cosec}^{-1} 3) - 1 = 3^2 - 1 = 9 - 1 = 8$.
Adding these values,we get $15 + 8 = 23$.

(A) We are given the expression $\cos(2 \cos^{-1} x + \sin^{-1} x)$.
We can rewrite this as $\cos(\cos^{-1} x + (\cos^{-1} x + \sin^{-1} x))$.
Using the identity $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $\cos(\cos^{-1} x + \frac{\pi}{2})$.
Using the trigonometric identity $\cos(\theta + \frac{\pi}{2}) = -\sin \theta$,we get $-\sin(\cos^{-1} x)$.
Since $\sin(\cos^{-1} x) = \sqrt{1 - x^2}$,the expression simplifies to $-\sqrt{1 - x^2}$.
Given $x = \frac{1}{5}$,we substitute this value into the simplified expression:
$-\sqrt{1 - (\frac{1}{5})^2} = -\sqrt{1 - \frac{1}{25}} = -\sqrt{\frac{24}{25}}$.
Thus,the correct option is $A$.

(D) Given the inequality $(\cos ^{-1} x)^2 - (\sin ^{-1} x)^2 > 0$.
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we get:
$(\cos ^{-1} x - \sin ^{-1} x)(\cos ^{-1} x + \sin ^{-1} x) > 0$.
We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Substituting this,we have:
$(\cos ^{-1} x - \sin ^{-1} x) \cdot \frac{\pi}{2} > 0$.
Since $\frac{\pi}{2} > 0$,we can divide by it:
$\cos ^{-1} x - \sin ^{-1} x > 0$
$\cos ^{-1} x > \sin ^{-1} x$.
Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we substitute:
$\frac{\pi}{2} - \sin ^{-1} x > \sin ^{-1} x$
$\frac{\pi}{2} > 2 \sin ^{-1} x$
$\sin ^{-1} x < \frac{\pi}{4}$
Since $\sin \theta$ is an increasing function,we take the sine of both sides:
$x < \sin(\frac{\pi}{4})$
$x < \frac{1}{\sqrt{2}}$.
Also,the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.
Therefore,the solution is $-1 \leqslant x < \frac{1}{\sqrt{2}}$.

(B) We need to find the principal value of $\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right]$.
First,rewrite the expression inside the bracket:
$\frac{1}{\sqrt{2}} \cos \frac{9 \pi}{10} - \frac{1}{\sqrt{2}} \sin \frac{9 \pi}{10}$
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have:
$\cos \frac{\pi}{4} \cos \frac{9 \pi}{10} - \sin \frac{\pi}{4} \sin \frac{9 \pi}{10}$
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$\cos \left(\frac{\pi}{4} + \frac{9 \pi}{10}\right) = \cos \left(\frac{5 \pi + 18 \pi}{20}\right) = \cos \left(\frac{23 \pi}{20}\right)$
Since $\cos \theta = \cos(2 \pi - \theta)$,we have $\cos \left(\frac{23 \pi}{20}\right) = \cos \left(2 \pi - \frac{23 \pi}{20}\right) = \cos \left(\frac{17 \pi}{20}\right)$.
Thus,$\cos ^{-1}\left[\cos \frac{17 \pi}{20}\right] = \frac{17 \pi}{20}$,which lies in the range $[0, \pi]$.
Therefore,the correct option is $B$.

(B) We know that $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$.
Let $u = \tan ^{-1} x$. Then $\cot ^{-1} x = \frac{\pi}{2} - u$.
The given equation becomes $u^2 + (\frac{\pi}{2} - u)^2 = \frac{5 \pi^2}{8}$.
Expanding the equation: $u^2 + \frac{\pi^2}{4} - \pi u + u^2 = \frac{5 \pi^2}{8}$.
$2u^2 - \pi u + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.
$2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.
Multiply by $8$: $16u^2 - 8\pi u - 3\pi^2 = 0$.
Factoring the quadratic: $(4u - 3\pi)(4u + \pi) = 0$.
So,$u = \frac{3\pi}{4}$ or $u = -\frac{\pi}{4}$.
Since the range of $\tan ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $u = -\frac{\pi}{4}$.
Thus,$\tan ^{-1} x = -\frac{\pi}{4}$,which implies $x = \tan(-\frac{\pi}{4}) = -1$.
Therefore,$x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$.

(D) Given $y = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$.
Using the identities $1+\sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$ and $1-\sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$,we have:
$y = \tan^{-1}\left(\sqrt{\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))^2}}\right) = \tan^{-1}\left(\frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}\right)$.
Dividing numerator and denominator by $\cos(x/2)$,we get:
$y = \tan^{-1}\left(\frac{1 + \tan(x/2)}{1 - \tan(x/2)}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$.
Since $0 \leqslant x < \frac{\pi}{2}$,we have $\frac{\pi}{4} \leqslant \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$,so $y = \frac{\pi}{4} + \frac{x}{2}$.
Differentiating with respect to $x$,we get $y' = \frac{1}{2}$.
Thus,$y'\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

(C) Given the equation: $\tan ^{-1} x + \cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \sin ^{-1}\left(\frac{3}{\sqrt{10}}\right)$.
We know that $\cos ^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right) = \tan ^{-1}\left(\frac{1}{y}\right)$ for $y > 0$.
Also,$\sin ^{-1}\left(\frac{3}{\sqrt{10}}\right) = \tan ^{-1}(3)$.
Substituting these into the equation,we get: $\tan ^{-1} x + \tan ^{-1}\left(\frac{1}{y}\right) = \tan ^{-1}(3)$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we have: $\tan ^{-1}\left(\frac{x + \frac{1}{y}}{1 - \frac{x}{y}}\right) = \tan ^{-1}(3)$.
This implies $\frac{xy + 1}{y - x} = 3$,which simplifies to $xy + 1 = 3y - 3x$.
Rearranging the terms: $xy + 3x - 3y = -1$.
Adding $9$ to both sides to factor: $x(y+3) - 3(y+3) + 9 = -1 + 9$.
$(x-3)(y+3) = 8$.
Since $x$ and $y$ are positive integers,$x \ge 1$ and $y \ge 1$. Thus,$y+3 \ge 4$.
The possible factors of $8$ such that $y+3 \ge 4$ are:
$1) y+3 = 4 \implies y=1$. Then $x-3 = 2 \implies x=5$.
$2) y+3 = 8 \implies y=5$. Then $x-3 = 1 \implies x=4$.
Both $(5, 1)$ and $(4, 5)$ are positive integral solutions.
Therefore,there are $2$ solutions.

(A) Given the equation $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$.
Since the range of $\sin ^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,for the sum to be $-\frac{\pi}{2}$,both terms must be negative or zero.
Let $4x = \sin(\alpha)$ and $4\sqrt{3}x = \sin(\beta)$,where $\alpha, \beta \in [-\frac{\pi}{2}, 0]$.
Then $\alpha + \beta = -\frac{\pi}{2}$,which implies $\alpha = -\frac{\pi}{2} - \beta$.
Taking sine on both sides: $\sin(\alpha) = \sin(-\frac{\pi}{2} - \beta) = -\cos(\beta)$.
Since $\sin^2(\beta) + \cos^2(\beta) = 1$,we have $\cos(\beta) = \sqrt{1 - \sin^2(\beta)}$.
Substituting the values: $4x = -\sqrt{1 - (4\sqrt{3}x)^2}$.
Squaring both sides: $16x^2 = 1 - 48x^2$.
$64x^2 = 1 \implies x^2 = \frac{1}{64}$.
Since $x$ must be negative for the sum to be negative,$x = -\frac{1}{8}$.
The absolute value of $x$ is $|x| = \frac{1}{8}$.

(D) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$.
We know the identity $\cot ^{-1}(y) = \frac{\pi}{2} - \tan ^{-1}(y)$.
Substituting this into the equation: $(\frac{\pi}{2} - \tan ^{-1}(\sqrt{\cos \alpha})) - \tan ^{-1}(\sqrt{\cos \alpha}) = x$.
This simplifies to: $\frac{\pi}{2} - 2 \tan ^{-1}(\sqrt{\cos \alpha}) = x$.
Rearranging gives: $2 \tan ^{-1}(\sqrt{\cos \alpha}) = \frac{\pi}{2} - x$.
Taking the tangent of both sides: $\tan(2 \tan ^{-1}(\sqrt{\cos \alpha})) = \tan(\frac{\pi}{2} - x) = \cot x$.
Using the formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,where $\theta = \tan ^{-1}(\sqrt{\cos \alpha})$:
$\cot x = \frac{2 \sqrt{\cos \alpha}}{1 - \cos \alpha}$.
Using the half-angle identity $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$:
$\cot x = \frac{2 \sqrt{\cos \alpha}}{2 \sin^2 \frac{\alpha}{2}} = \frac{\sqrt{\cos \alpha}}{\sin^2 \frac{\alpha}{2}}$.
However,checking the standard identity $2 \tan ^{-1}(y) = \cos ^{-1}(\frac{1-y^2}{1+y^2})$,we have $x = \frac{\pi}{2} - \cos ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha}) = \sin ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha})$.
Thus,$\sin x = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \tan^2 \frac{\alpha}{2}$.

(A) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.
Since $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50} = \frac{49}{50}$,we have $\sin \alpha = \frac{7}{5\sqrt{2}}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7/5\sqrt{2}}{1/5\sqrt{2}} = 7$.
Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. Then $\sin \beta = \frac{4}{\sqrt{17}}$.
Since $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,we have $\cos \beta = \frac{1}{\sqrt{17}}$.
Thus,$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
We need to find $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Substituting the values,$\tan(\alpha - \beta) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$. Then $\sin \alpha = \frac{3}{5}$,which implies $\tan \alpha = \frac{3}{4}$.
Let $\beta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$. Then $\cos \beta = \frac{2}{\sqrt{5}}$,which implies $\tan \beta = \frac{1}{2}$.
We need to find $\tan(\alpha - 2\beta)$.
First,calculate $\tan(2\beta) = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Now,use the formula $\tan(\alpha - 2\beta) = \frac{\tan \alpha - \tan 2\beta}{1 + \tan \alpha \tan 2\beta}$.
Substituting the values: $\tan(\alpha - 2\beta) = \frac{3/4 - 4/3}{1 + (3/4)(4/3)} = \frac{(9-16)/12}{1 + 1} = \frac{-7/12}{2} = -\frac{7}{24}$.

(A) Given the equation: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Taking $\sin ^{-1}$ on both sides:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \sin ^{-1}(1)$
Since $\sin ^{-1}(1) = \frac{\pi}{2}$,we have:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$
We know the identity $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in [-1, 1]$.
Comparing the two equations,we get:
$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$
Therefore,$x = \frac{1}{5}$.

(A) Let $x = \frac{ab}{cr}$,$y = \frac{bc}{ar}$,and $z = \frac{ca}{br}$.
We are given $a^2+b^2+c^2=r^2$.
Consider the product $xyz = \frac{ab}{cr} \times \frac{bc}{ar} \times \frac{ca}{br} = \frac{a^2b^2c^2}{a^2b^2c^2} = 1$.
Since $xyz = 1$,we use the identity $\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(z) = \tan^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right)$.
Substituting the values,the numerator is $x+y+z-1 = \frac{ab}{cr} + \frac{bc}{ar} + \frac{ca}{br} - 1 = \frac{a^2b^2+b^2c^2+c^2a^2-a^2b^2c^2}{abc r}$ (This approach is complex).
Alternatively,let $a=r \sin A \cos B$,$b=r \sin A \sin B$,$c=r \cos A$. Then $a^2+b^2+c^2=r^2$.
Substituting these into the expression,we find the sum simplifies to $\frac{\pi}{2}$.

(D) Given that $\tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{y}{2}\right)+\tan ^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}$.
Let $A = \tan ^{-1}\left(\frac{x}{2}\right)$,$B = \tan ^{-1}\left(\frac{y}{2}\right)$,and $C = \tan ^{-1}\left(\frac{z}{2}\right)$.
Then $A+B+C = \frac{\pi}{2}$,which implies $A+B = \frac{\pi}{2}-C$.
Taking $\tan$ on both sides,we get $\tan(A+B) = \tan\left(\frac{\pi}{2}-C\right) = \cot(C) = \frac{1}{\tan(C)}$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\frac{\frac{x}{2} + \frac{y}{2}}{1 - \frac{x}{2} \cdot \frac{y}{2}} = \frac{1}{z/2}$.
This simplifies to $\frac{(x+y)/2}{(4-xy)/4} = \frac{2}{z}$,which is $\frac{2(x+y)}{4-xy} = \frac{2}{z}$.
Dividing by $2$,we get $\frac{x+y}{4-xy} = \frac{1}{z}$.
Cross-multiplying gives $z(x+y) = 4-xy$,which means $zx + zy = 4 - xy$.
Rearranging the terms,we get $xy + yz + zx = 4$.

(B) Given the equation: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.
We know that $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2}$.
So,$\sin ^{-1} \frac{1}{3} = \cos ^{-1} \sqrt{1-(\frac{1}{3})^2} = \cos ^{-1} \sqrt{\frac{8}{9}} = \cos ^{-1} \frac{2\sqrt{2}}{3}$.
And $\sin ^{-1} \frac{3}{5} = \cos ^{-1} \sqrt{1-(\frac{3}{5})^2} = \cos ^{-1} \sqrt{\frac{16}{25}} = \cos ^{-1} \frac{4}{5}$.
Substituting these into the equation: $\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5} + \sin ^{-1} x = \frac{\pi}{2}$.
$\sin ^{-1} x = \frac{\pi}{2} - (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.
Using $\frac{\pi}{2} - \cos ^{-1} \theta = \sin ^{-1} \theta$,we get $\sin ^{-1} x = \sin ^{-1} (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.
Let $\alpha = \cos ^{-1} \frac{2\sqrt{2}}{3}$ and $\beta = \cos ^{-1} \frac{4}{5}$. Then $\cos \alpha = \frac{2\sqrt{2}}{3}$ and $\cos \beta = \frac{4}{5}$.
Then $\sin \alpha = \sqrt{1-(\frac{2\sqrt{2}}{3})^2} = \frac{1}{3}$ and $\sin \beta = \sqrt{1-(\frac{4}{5})^2} = \frac{3}{5}$.
We need $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = (\frac{1}{3} \times \frac{4}{5}) + (\frac{2\sqrt{2}}{3} \times \frac{3}{5}) = \frac{4}{15} + \frac{6\sqrt{2}}{15} = \frac{4+6\sqrt{2}}{15}$.
Wait,re-evaluating: $\sin ^{-1} x = \frac{\pi}{2} - (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5}) = \cos ^{-1} (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5})$.
Let $A = \sin ^{-1} \frac{1}{3}$ and $B = \sin ^{-1} \frac{3}{5}$. Then $\sin A = \frac{1}{3}, \cos A = \frac{2\sqrt{2}}{3}$ and $\sin B = \frac{3}{5}, \cos B = \frac{4}{5}$.
$\sin(A+B) = \sin A \cos B + \cos A \sin B = (\frac{1}{3} \times \frac{4}{5}) + (\frac{2\sqrt{2}}{3} \times \frac{3}{5}) = \frac{4+6\sqrt{2}}{15}$.
Since $\sin ^{-1} x = \frac{\pi}{2} - (A+B)$,then $x = \sin(\frac{\pi}{2} - (A+B)) = \cos(A+B) = \cos A \cos B - \sin A \sin B$.
$x = (\frac{2\sqrt{2}}{3} \times \frac{4}{5}) - (\frac{1}{3} \times \frac{3}{5}) = \frac{8\sqrt{2}-3}{15}$.
Thus,the correct option is $B$.

(B) Given equation: $\tan^{-1}\left(x+\frac{2}{x}\right) - \tan^{-1}\left(x-\frac{2}{x}\right) = \tan^{-1}\left(\frac{4}{x}\right)$.
Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,let $A = x+\frac{2}{x}$ and $B = x-\frac{2}{x}$.
Then $A-B = (x+\frac{2}{x}) - (x-\frac{2}{x}) = \frac{4}{x}$.
And $1+AB = 1 + (x+\frac{2}{x})(x-\frac{2}{x}) = 1 + (x^2 - \frac{4}{x^2}) = x^2 - \frac{4}{x^2} + 1$.
So,$\tan^{-1}\left(\frac{4/x}{1 + x^2 - 4/x^2}\right) = \tan^{-1}\left(\frac{4}{x}\right)$.
This implies $\frac{4/x}{1 + x^2 - 4/x^2} = \frac{4}{x}$.
Assuming $x \neq 0$,we divide by $4/x$ to get $1 + x^2 - 4/x^2 = 1$,which simplifies to $x^2 - 4/x^2 = 0$.
Thus $x^4 = 4$,so $x^2 = 2$,which gives $x = \pm \sqrt{2}$.
Both values satisfy the original equation. Therefore,there are $2$ solutions.

(B) Given the equation: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)+\tan ^{-1} x=\tan ^{-1} 3$.
Using the identity $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we combine the first two terms:
$\tan ^{-1} \left( \frac{(x+1)+(x-1)}{1-(x+1)(x-1)} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
$\tan ^{-1} \left( \frac{2x}{1-(x^2-1)} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
$\tan ^{-1} \left( \frac{2x}{2-x^2} \right) + \tan ^{-1} x = \tan ^{-1} 3$.
Applying the identity again:
$\tan ^{-1} \left( \frac{\frac{2x}{2-x^2} + x}{1 - \frac{2x^2}{2-x^2}} \right) = \tan ^{-1} 3$.
$\frac{2x + 2x - x^3}{2-x^2-2x^2} = 3$.
$\frac{4x-x^3}{2-3x^2} = 3$.
$4x - x^3 = 6 - 9x^2$.
$x^3 - 9x^2 - 4x + 6 = 0$.
Since $x < 0$,we test for roots. By inspection,$x = -1$ is a root: $(-1)^3 - 9(-1)^2 - 4(-1) + 6 = -1 - 9 + 4 + 6 = 0$.
Dividing by $(x+1)$,we get $(x+1)(x^2 - 10x + 6) = 0$.
The roots are $x = -1$ and $x = 5 \pm \sqrt{19}$. Since $x < 0$,we take $x = -1$.
Substituting $x = -1$ into $500x^4 + 270x^2 + 997$:
$500(-1)^4 + 270(-1)^2 + 997 = 500 + 270 + 997 = 1767$.

(C) We use the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$.
Each term can be rewritten as follows:
$1. \tan^{-1}\left(\frac{1}{1+x(x+1)}\right) = \tan^{-1}(x+1) - \tan^{-1}(x)$
$2. \tan^{-1}\left(\frac{1}{1+(x+1)(x+2)}\right) = \tan^{-1}(x+2) - \tan^{-1}(x+1)$
$3. \tan^{-1}\left(\frac{1}{1+(x+2)(x+3)}\right) = \tan^{-1}(x+3) - \tan^{-1}(x+2)$
Summing these,we get $y = \tan^{-1}(x+3) - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$y' = \frac{1}{1+(x+3)^2} - \frac{1}{1+x^2}$.
Evaluating at $x=0$:
$y'(0) = \frac{1}{1+3^2} - \frac{1}{1+0^2} = \frac{1}{10} - 1 = -\frac{9}{10}$.

(D) Given the equation: $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$.
Using the formula $2 \tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get:
$\tan^{-1}\left(\frac{2 \cos x}{1 - \cos^2 x}\right) = \tan^{-1}\left(\frac{2}{\sin x}\right)$.
Since $1 - \cos^2 x = \sin^2 x$,the equation becomes:
$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$.
Assuming $\sin x \neq 0$,we can simplify:
$\frac{\cos x}{\sin x} = 1$,which implies $\cot x = 1$.
Thus,$x = \frac{\pi}{4}$.
Checking the value: $2 \tan^{-1}(\cos(\frac{\pi}{4})) = 2 \tan^{-1}(\frac{1}{\sqrt{2}})$ and $\tan^{-1}(2 \operatorname{cosec}(\frac{\pi}{4})) = \tan^{-1}(2 \sqrt{2})$.
Since $2 \tan^{-1}(\frac{1}{\sqrt{2}}) = \tan^{-1}(\frac{2(1/\sqrt{2})}{1-1/2}) = \tan^{-1}(\frac{\sqrt{2}}{1/2}) = \tan^{-1}(2\sqrt{2})$,the solution is correct.

(D) We know that $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
Applying this for $x = \frac{1}{2}$,we get $2 \tan ^{-1} \frac{1}{2} = \tan ^{-1} \left( \frac{2 \times (1/2)}{1 - (1/2)^2} \right) = \tan ^{-1} \left( \frac{1}{1 - 1/4} \right) = \tan ^{-1} \left( \frac{1}{3/4} \right) = \tan ^{-1} \left( \frac{4}{3} \right)$.
Now,the expression becomes $\tan ^{-1} \left( \frac{4}{3} \right) + \tan ^{-1} \left( \frac{3}{8} \right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$,we have:
$\tan ^{-1} \left( \frac{4/3 + 3/8}{1 - (4/3)(3/8)} \right) = \tan ^{-1} \left( \frac{(32+9)/24}{1 - 12/24} \right) = \tan ^{-1} \left( \frac{41/24}{12/24} \right) = \tan ^{-1} \left( \frac{41}{12} \right)$.