Mix Examples-Trigonometrical Ratios, Functions and Identities Questions in English

(C) Given the equation: $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$
Multiply both sides by $\cos x(1 + \cos x)$:
$1 - \cos x = \sin \alpha (1 + \cos x) - 2 \cos x$
$1 - \cos x = \sin \alpha + \sin \alpha \cos x - 2 \cos x$
$1 - \cos x = \sin \alpha + \cos x(\sin \alpha - 2)$
Comparing the constant terms and the coefficients of $\cos x$ on both sides:
Constant term: $1 = \sin \alpha$
Coefficient of $\cos x$: $-1 = \sin \alpha - 2$
From the constant term,$\sin \alpha = 1$,which implies $\alpha = \frac{\pi}{2}$.
Checking with the coefficient: $-1 = 1 - 2$,which is $-1 = -1$.
Thus,$\alpha = \frac{\pi}{2}$.

(A) Let the given expression be $S = \sum \frac{1}{1 + x^{a - b} + x^{a - c}}$.
Multiplying the numerator and denominator of each term by $x^a$,we get:
$S = \sum \frac{x^a}{x^a + x^b + x^c}$.
Wait,the standard form for this cyclic sum is $\sum \frac{1}{1 + x^{b-a} + x^{c-a}} = \sum \frac{x^a}{x^a + x^b + x^c} = 1$ if $a+b+c=0$ or similar constraints.
Given the expression $\sum \frac{1}{1 + x^{a-b} + x^{a-c}} = \sum \frac{x^b x^c}{x^b x^c + x^a x^c + x^a x^b} = \sum \frac{x^{b+c}}{x^{b+c} + x^{a+c} + x^{a+b}}$.
Since the denominator $x^{a+b} + x^{b+c} + x^{c+a}$ is common for all terms in the cyclic sum:
$S = \frac{1}{x^{a+b} + x^{b+c} + x^{c+a}} \sum (x^{b+c} + x^{a+c} + x^{a+b}) = 1$.

(C) Given: $\cos \alpha + \cos \beta + \cos \gamma = 0$ $(i)$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ $(ii)$.
Let $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.
Then $a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 + i(0) = 0$.
Since $a, b, c$ are complex numbers on the unit circle,$|a| = |b| = |c| = 1$,so $\bar{a} = 1/a$,$\bar{b} = 1/b$,and $\bar{c} = 1/c$.
Taking the conjugate of $a + b + c = 0$,we get $\bar{a} + \bar{b} + \bar{c} = 0$,which implies $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$.
This simplifies to $\frac{ab + bc + ca}{abc} = 0$,so $ab + bc + ca = 0$.
Now,consider $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$.
Since $ab + bc + ca = 0$,we have $a^2 + b^2 + c^2 = 0$.
Substituting $a^2 = e^{i2\alpha} = \cos 2\alpha + i\sin 2\alpha$,etc.,we get:
$(\cos 2\alpha + \cos 2\beta + \cos 2\gamma) + i(\sin 2\alpha + \sin 2\beta + \sin 2\gamma) = 0$.
Equating the real parts,we get $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$.

(D) Given equation: $e^{\sin x} - e^{-\sin x} - 4 = 0$
Let $y = e^{\sin x}$. Since $e^{-\sin x} = \frac{1}{y}$,the equation becomes:
$y - \frac{1}{y} - 4 = 0$
Multiplying by $y$ $(y \neq 0)$:
$y^2 - 4y - 1 = 0$
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$
Since $y = e^{\sin x} > 0$,we must have $y = 2 + \sqrt{5}$ (as $2 - \sqrt{5} < 0$).
Taking the natural logarithm on both sides:
$\sin x = \ln(2 + \sqrt{5})$
We know that $\sqrt{4} < \sqrt{5} < \sqrt{9}$,so $2 < \sqrt{5} < 3$. Thus,$4 < 2 + \sqrt{5} < 5$.
Since $e^1 \approx 2.718$,$\ln(2 + \sqrt{5}) > \ln(e) = 1$.
Since $\sin x \leq 1$ for all real $x$,the equation $\sin x = \ln(2 + \sqrt{5})$ has no real solution.
Therefore,the number of real roots is $0$ (None).

(D) Given $\sin \theta + \text{cosec} \theta = 2$.
Since $\text{cosec} \theta = \frac{1}{\sin \theta}$,we have $\sin \theta + \frac{1}{\sin \theta} = 2$.
Multiplying by $\sin \theta$,we get $\sin^2 \theta + 1 = 2 \sin \theta$.
This simplifies to $\sin^2 \theta - 2 \sin \theta + 1 = 0$,which is $(\sin \theta - 1)^2 = 0$.
Thus,$\sin \theta = 1$.
Therefore,$\sin^{10} \theta + \text{cosec}^{10} \theta = (1)^{10} + \frac{1}{(1)^{10}} = 1 + 1 = 2$.

(C) Given that $\sin \theta + \text{cosec} \theta = 2$.
Squaring both sides,we get:
$(\sin \theta + \text{cosec} \theta)^2 = 2^2$
$\sin^2 \theta + \text{cosec}^2 \theta + 2 \sin \theta \text{cosec} \theta = 4$
Since $\sin \theta \text{cosec} \theta = 1$,the equation becomes:
$\sin^2 \theta + \text{cosec}^2 \theta + 2(1) = 4$
$\sin^2 \theta + \text{cosec}^2 \theta = 4 - 2 = 2$
Thus,the correct option is $C$.

(C) Given: $\sin \theta + \cos \theta = m$ and $\sec \theta + \text{cosec} \theta = n$.
Squaring the first equation: $(\sin \theta + \cos \theta)^2 = m^2 \implies 1 + 2 \sin \theta \cos \theta = m^2 \implies 2 \sin \theta \cos \theta = m^2 - 1$.
Now,consider the expression $n(m + 1)(m - 1) = n(m^2 - 1)$.
Substitute $n = \sec \theta + \text{cosec} \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{m}{\sin \theta \cos \theta}$.
Thus,$n(m^2 - 1) = \left( \frac{m}{\sin \theta \cos \theta} \right) (2 \sin \theta \cos \theta) = 2m$.

(A) Given: $\sin \theta + \cos \theta = 1$
Squaring both sides:
$(\sin \theta + \cos \theta)^2 = 1^2$
Using the identity $(\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cos \theta = 1$:
$1 + 2 \sin \theta \cos \theta = 1$
Subtracting $1$ from both sides:
$2 \sin \theta \cos \theta = 0$
Dividing by $2$:
$\sin \theta \cos \theta = 0$

(C) We know that $\text{cosec}^2 A - \cot^2 A = 1.$
Using the identity $a^2 - b^2 = (a - b)(a + b),$ we have $(\text{cosec } A - \cot A)(\text{cosec } A + \cot A) = 1.$
Given $\text{cosec } A + \cot A = \frac{11}{2},$ we substitute this into the identity:
$(\text{cosec } A - \cot A) \times \frac{11}{2} = 1 \Rightarrow \text{cosec } A - \cot A = \frac{2}{11}.$
Now,subtract the two equations:
$(\text{cosec } A + \cot A) - (\text{cosec } A - \cot A) = \frac{11}{2} - \frac{2}{11}.$
$2 \cot A = \frac{121 - 4}{22} = \frac{117}{22}.$
$\cot A = \frac{117}{44}.$
Since $\tan A = \frac{1}{\cot A},$ we get $\tan A = \frac{44}{117}.$

(B) Given equation: $(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1$.
Divide by $\cos \theta$ (assuming $\cos \theta \neq 0$): $(m + 2)\tan \theta + (2m - 1) = (2m + 1)\sec \theta$.
Squaring both sides: $((m + 2)\tan \theta + (2m - 1))^2 = (2m + 1)^2(1 + \tan^2 \theta)$.
Let $\tan \theta = t$. Then $(m + 2)^2 t^2 + 2(m + 2)(2m - 1)t + (2m - 1)^2 = (2m + 1)^2(1 + t^2)$.
Expanding and simplifying: $(m^2 + 4m + 4)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1) = (4m^2 + 4m + 1)(1 + t^2)$.
$(m^2 + 4m + 4)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1) = (4m^2 + 4m + 1) + (4m^2 + 4m + 1)t^2$.
Rearranging terms: $(3m^2 - 3)t^2 - (4m^2 + 6m - 4)t + 4m = 0$.
$(3t - 4)((m^2 - 1)t - 2m) = 0$.
Thus,$t = \frac{4}{3}$ or $t = \frac{2m}{m^2 - 1}$.
Therefore,$\tan \theta = \frac{4}{3}$ is a valid solution.

(B) Given: $\sin x + \sin y = 3(\cos y - \cos x)$
Rearranging the terms,we get: $\sin x + 3\cos x = 3\cos y - \sin y$.....$(i)$
Divide both sides by $\sqrt{1^2 + 3^2} = \sqrt{10}$:
$\frac{1}{\sqrt{10}}\sin x + \frac{3}{\sqrt{10}}\cos x = \frac{3}{\sqrt{10}}\cos y - \frac{1}{\sqrt{10}}\sin y$
Let $\cos \alpha = \frac{1}{\sqrt{10}}$ and $\sin \alpha = \frac{3}{\sqrt{10}}$,so $\tan \alpha = 3$. Then the equation becomes:
$\sin(x + \alpha) = \sin(\alpha - y)$
This implies $x + \alpha = n\pi + (-1)^n(\alpha - y)$.
For $n=0$,$x + \alpha = \alpha - y \Rightarrow x = -y$.
Substituting $x = -y$ into the expression $\frac{\sin 3x}{\sin 3y}$:
$\frac{\sin 3(-y)}{\sin 3y} = \frac{-\sin 3y}{\sin 3y} = -1$.

(A) Given that $\sin A, \cos A, \tan A$ are in $G.P.$
Therefore,the square of the middle term is equal to the product of the extremes:
$\cos^2 A = \sin A \cdot \tan A$
Substitute $\tan A = \frac{\sin A}{\cos A}$:
$\cos^2 A = \sin A \cdot \frac{\sin A}{\cos A}$
$\cos^3 A = \sin^2 A$
Using the identity $\sin^2 A = 1 - \cos^2 A$:
$\cos^3 A = 1 - \cos^2 A$
Rearranging the terms:
$\cos^3 A + \cos^2 A = 1$

(B) Given that $\tan \theta + \sec \theta = e^x$ $(i)$.
We know the identity $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting $(i)$ into this identity,we get $\sec \theta - \tan \theta = \frac{1}{e^x} = e^{-x}$ $(ii)$.
Adding equations $(i)$ and $(ii)$:
$(\tan \theta + \sec \theta) + (\sec \theta - \tan \theta) = e^x + e^{-x}$.
$2 \sec \theta = e^x + e^{-x}$.
Since $\cos \theta = \frac{1}{\sec \theta}$,we have $\cos \theta = \frac{2}{e^x + e^{-x}}$.

(A) Given that $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$.
Rearranging the terms,we get $\cos \theta = (\sqrt{2} + 1) \sin \theta$.
To isolate $\sin \theta$,multiply both sides by $(\sqrt{2} - 1)$:
$(\sqrt{2} - 1) \cos \theta = (\sqrt{2} - 1)(\sqrt{2} + 1) \sin \theta$.
Since $(\sqrt{2} - 1)(\sqrt{2} + 1) = 2 - 1 = 1$,we have:
$(\sqrt{2} - 1) \cos \theta = \sin \theta$.
Expanding this,$\sqrt{2} \cos \theta - \cos \theta = \sin \theta$.
Therefore,$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$.

(B) Given that $\sec \theta + \tan \theta = p$ $(i)$
We know that $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting $(i)$ into this identity,we get $\sec \theta - \tan \theta = \frac{1}{p}$ $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = p - \frac{1}{p}$
$2 \tan \theta = \frac{p^2 - 1}{p}$
$\tan \theta = \frac{p^2 - 1}{2p}$.

(D) Let the given expression be $X = e^{\log_{10} \tan 1^\circ + \log_{10} \tan 2^\circ + \dots + \log_{10} \tan 89^\circ}$.
Using the property $\log a + \log b = \log(ab)$,we get:
$X = e^{\log_{10}(\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \dots \cdot \tan 89^\circ)}$.
We know that $\tan(90^\circ - \theta) = \cot \theta$,so $\tan 89^\circ = \cot 1^\circ$,$\tan 88^\circ = \cot 2^\circ$,and so on.
The product $P = \tan 1^\circ \cdot \tan 2^\circ \cdot \dots \cdot \tan 44^\circ \cdot \tan 45^\circ \cdot \tan 46^\circ \cdot \dots \cdot \tan 89^\circ$.
$P = (\tan 1^\circ \cdot \cot 1^\circ) \cdot (\tan 2^\circ \cdot \cot 2^\circ) \cdot \dots \cdot (\tan 44^\circ \cdot \cot 44^\circ) \cdot \tan 45^\circ$.
Since $\tan \theta \cdot \cot \theta = 1$ and $\tan 45^\circ = 1$,we have $P = 1 \cdot 1 \cdot \dots \cdot 1 = 1$.
Thus,$X = e^{\log_{10}(1)} = e^0 = 1$.

(B) Given expression:
$E = \frac{2\sin \theta \tan \theta (1 - \tan \theta) + 2\sin \theta \sec^2 \theta}{(1 + \tan \theta)^2}$
Factor out $2\sin \theta$:
$E = \frac{2\sin \theta [\tan \theta (1 - \tan \theta) + \sec^2 \theta]}{(1 + \tan \theta)^2}$
Expand the terms inside the bracket:
$E = \frac{2\sin \theta [\tan \theta - \tan^2 \theta + 1 + \tan^2 \theta]}{(1 + \tan \theta)^2}$
Simplify the terms inside the bracket:
$E = \frac{2\sin \theta [1 + \tan \theta]}{(1 + \tan \theta)^2}$
Cancel $(1 + \tan \theta)$:
$E = \frac{2\sin \theta}{1 + \tan \theta}$

(D) Given expression: $E = 1 - \frac{\sin^2 y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y}$
Step $1$: Simplify the first two terms:
$1 - \frac{\sin^2 y}{1 + \cos y} = \frac{1 + \cos y - (1 - \cos^2 y)}{1 + \cos y} = \frac{\cos y + \cos^2 y}{1 + \cos y} = \frac{\cos y(1 + \cos y)}{1 + \cos y} = \cos y$
Step $2$: Simplify the last two terms:
$\frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y} = \frac{(1 + \cos y)(1 - \cos y) - \sin^2 y}{\sin y(1 - \cos y)} = \frac{1 - \cos^2 y - \sin^2 y}{\sin y(1 - \cos y)} = \frac{\sin^2 y - \sin^2 y}{\sin y(1 - \cos y)} = 0$
Step $3$: Combine the results:
$E = \cos y + 0 = \cos y$

(A) Given equations are:
$2y \cos \theta = x \sin \theta$ --- $(i)$
$2x \sec \theta - y \csc \theta = 3$ --- $(ii)$
From $(i)$,we have $\frac{x}{\cos \theta} = \frac{2y}{\sin \theta} = k$ (let).
So,$x = k \cos \theta$ and $2y = k \sin \theta$,which implies $y = \frac{k}{2} \sin \theta$.
Substitute these into $(ii)$:
$2(k \cos \theta) \sec \theta - (\frac{k}{2} \sin \theta) \csc \theta = 3$
$2k(1) - \frac{k}{2}(1) = 3$
$\frac{3k}{2} = 3 \Rightarrow k = 2$.
Thus,$x = 2 \cos \theta$ and $y = \sin \theta$.
Now,$x^2 + 4y^2 = (2 \cos \theta)^2 + 4(\sin \theta)^2$
$= 4 \cos^2 \theta + 4 \sin^2 \theta$
$= 4(\cos^2 \theta + \sin^2 \theta) = 4(1) = 4$.

(B) Given $x = \sec \phi - \tan \phi = \frac{1 - \sin \phi}{\cos \phi}$ and $y = \csc \phi + \cot \phi = \frac{1 + \cos \phi}{\sin \phi}$.
Consider the expression $\frac{y - 1}{y + 1}$:
$y - 1 = \frac{1 + \cos \phi}{\sin \phi} - 1 = \frac{1 + \cos \phi - \sin \phi}{\sin \phi}$
$y + 1 = \frac{1 + \cos \phi}{\sin \phi} + 1 = \frac{1 + \cos \phi + \sin \phi}{\sin \phi}$
Therefore,$\frac{y - 1}{y + 1} = \frac{1 + \cos \phi - \sin \phi}{1 + \cos \phi + \sin \phi}$.
Multiply the numerator and denominator by $(1 + \cos \phi - \sin \phi)$:
$= \frac{(1 + \cos \phi - \sin \phi)^2}{(1 + \cos \phi)^2 - \sin^2 \phi} = \frac{1 + \cos^2 \phi + \sin^2 \phi + 2\cos \phi - 2\sin \phi - 2\sin \phi \cos \phi}{1 + \cos^2 \phi + 2\cos \phi - \sin^2 \phi}$
$= \frac{2 + 2\cos \phi - 2\sin \phi - 2\sin \phi \cos \phi}{2\cos^2 \phi + 2\cos \phi} = \frac{2(1 + \cos \phi)(1 - \sin \phi)}{2\cos \phi(1 + \cos \phi)}$
$= \frac{1 - \sin \phi}{\cos \phi} = x$.
Thus,$x = \frac{y - 1}{y + 1}$.

(B) Given $\tan \theta = \frac{x \sin \phi}{1 - x \cos \phi}$.
Rearranging,we get $\tan \theta - x \tan \theta \cos \phi = x \sin \phi$.
$\tan \theta = x(\sin \phi + \cos \phi \tan \theta) = x \left( \frac{\sin \phi \cos \theta + \cos \phi \sin \theta}{\cos \theta} \right) = x \frac{\sin(\phi + \theta)}{\cos \theta}$.
Thus,$x = \frac{\tan \theta \cos \theta}{\sin(\theta + \phi)} = \frac{\sin \theta}{\sin(\theta + \phi)}$.
Similarly,from $\tan \phi = \frac{y \sin \theta}{1 - y \cos \theta}$,we get $y = \frac{\sin \phi}{\sin(\theta + \phi)}$.
Therefore,$\frac{x}{y} = \frac{\sin \theta / \sin(\theta + \phi)}{\sin \phi / \sin(\theta + \phi)} = \frac{\sin \theta}{\sin \phi}$.

(D) Given $p = \frac{2\sin \theta}{1 + \cos \theta + \sin \theta}$ and $q = \frac{\cos \theta}{1 + \sin \theta}$.
Consider $p + q = \frac{2\sin \theta}{1 + \sin \theta + \cos \theta} + \frac{\cos \theta}{1 + \sin \theta}$.
$p + q = \frac{2\sin \theta (1 + \sin \theta) + \cos \theta (1 + \sin \theta + \cos \theta)}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)}$.
$p + q = \frac{2\sin \theta + 2\sin^2 \theta + \cos \theta + \sin \theta \cos \theta + \cos^2 \theta}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$p + q = \frac{2\sin \theta + \sin^2 \theta + 1 + \cos \theta + \sin \theta \cos \theta}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)}$.
$p + q = \frac{\sin^2 \theta + 2\sin \theta + 1 + \cos \theta (1 + \sin \theta)}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)}$.
$p + q = \frac{(1 + \sin \theta)^2 + \cos \theta (1 + \sin \theta)}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)}$.
$p + q = \frac{(1 + \sin \theta)(1 + \sin \theta + \cos \theta)}{(1 + \sin \theta + \cos \theta)(1 + \sin \theta)} = 1$.

(D) Given: $m = \tan \theta + \sin \theta$ and $n = \tan \theta - \sin \theta$.
Calculate $m^2 - n^2$:
$m^2 - n^2 = (m + n)(m - n)$
$m + n = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = 2 \tan \theta$
$m - n = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = 2 \sin \theta$
$m^2 - n^2 = (2 \tan \theta)(2 \sin \theta) = 4 \tan \theta \sin \theta$ ... $(i)$
Calculate $4\sqrt{mn}$:
$mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta$
$mn = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \frac{\sin^2 \theta}{\cos^2 \theta} = \sin^2 \theta \tan^2 \theta$
$\sqrt{mn} = \sqrt{\sin^2 \theta \tan^2 \theta} = \sin \theta \tan \theta$
$4\sqrt{mn} = 4 \sin \theta \tan \theta$ ... $(ii)$
From $(i)$ and $(ii)$,we get ${m^2} - {n^2} = 4\sqrt {mn}$.

(C) Given that $a \cos \theta + b \sin \theta = m$ $(1)$
and $a \sin \theta - b \cos \theta = n$ $(2)$
Squaring both equations and adding them,we get:
$(a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = m^2 + n^2$
Expanding the squares:
$(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta) = m^2 + n^2$
Grouping the terms:
$a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = m^2 + n^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$a^2(1) + b^2(1) = m^2 + n^2$
Therefore,$a^2 + b^2 = m^2 + n^2$.

(C) Given $x = a \cos^3 \theta$ and $y = b \sin^3 \theta$.
Dividing by $a$ and $b$ respectively,we get $\frac{x}{a} = \cos^3 \theta$ and $\frac{y}{b} = \sin^3 \theta$.
Taking the power of $1/3$ on both sides,we get $(\frac{x}{a})^{1/3} = \cos \theta$ and $(\frac{y}{b})^{1/3} = \sin \theta$.
Squaring both sides,we get $(\frac{x}{a})^{2/3} = \cos^2 \theta$ and $(\frac{y}{b})^{2/3} = \sin^2 \theta$.
Adding these two equations,we get $(\frac{x}{a})^{2/3} + (\frac{y}{b})^{2/3} = \cos^2 \theta + \sin^2 \theta$.
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have $(\frac{x}{a})^{2/3} + (\frac{y}{b})^{2/3} = 1$.

(A) Given $\cot \theta + \tan \theta = m$
$\Rightarrow \frac{1}{\tan \theta} + \tan \theta = m$ $\Rightarrow \frac{1 + \tan^2 \theta}{\tan \theta} = m$ $\Rightarrow \sec^2 \theta = m \tan \theta$ ... $(i)$
Given $\sec \theta - \cos \theta = n$
$\Rightarrow \frac{1}{\cos \theta} - \cos \theta = n$ $\Rightarrow \frac{1 - \cos^2 \theta}{\cos \theta} = n$ $\Rightarrow \sin^2 \theta = n \cos \theta$
$\Rightarrow \tan^2 \theta \cos^2 \theta = n \cos \theta$ $\Rightarrow \tan^2 \theta \cos \theta = n$ $\Rightarrow \tan^2 \theta = n \sec \theta$ ... $(ii)$
From $(i)$,$\tan \theta = \frac{\sec^2 \theta}{m}$. Substituting in $(ii)$:
$(\frac{\sec^2 \theta}{m})^2 = n \sec \theta$ $\Rightarrow \frac{\sec^4 \theta}{m^2} = n \sec \theta$ $\Rightarrow \sec^3 \theta = m^2 n$ $\Rightarrow \sec \theta = (m^2 n)^{1/3}$
From $(i)$,$\tan \theta = \frac{\sec^2 \theta}{m} = \frac{(m^2 n)^{2/3}}{m} = (m n^2)^{1/3}$
Using $\sec^2 \theta - \tan^2 \theta = 1$:
$(m^2 n)^{2/3} - (m n^2)^{2/3} = 1$
Multiplying by $m/m$ and $n/n$ respectively:
$m \frac{(m^2 n)^{2/3}}{m} - n \frac{(m n^2)^{2/3}}{n} = 1$
$m (m n^2)^{1/3} - n (n m^2)^{1/3} = 1$.

(C) We know the algebraic identity $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$.
Let $a = \sin^2 \theta$ and $b = \cos^2 \theta$.
Then,$\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 = (\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta)$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin^6 \theta + \cos^6 \theta = (1)^3 - 3 \sin^2 \theta \cos^2 \theta (1) = 1 - 3 \sin^2 \theta \cos^2 \theta$.
Substituting this into the original expression:
$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = (1 - 3 \sin^2 \theta \cos^2 \theta) + 3 \sin^2 \theta \cos^2 \theta = 1$.
Alternatively,by putting $\theta = 0^\circ$,we get $\sin^6(0^\circ) + \cos^6(0^\circ) + 3 \sin^2(0^\circ) \cos^2(0^\circ) = 0 + 1 + 0 = 1$.

(B) We know that $\sin^2 \theta + \cos^2 \theta = 1$.
First,consider $(\sin^2 \theta + \cos^2 \theta)^3 = 1^3$,which expands to $\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin^6 \theta + \cos^6 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta$.
Next,consider $(\sin^2 \theta + \cos^2 \theta)^2 = 1^2$,which expands to $\sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta = 1$.
Thus,$\sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting these into the expression:
$2(1 - 3 \sin^2 \theta \cos^2 \theta) - 3(1 - 2 \sin^2 \theta \cos^2 \theta) + 1$
$= 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1$
$= 2 - 3 + 1 = 0$.

(C) Given,$\sin x + \sin^2 x = 1$.
This implies $\sin x = 1 - \sin^2 x$,so $\sin x = \cos^2 x$.
Now,consider the expression: $E = \cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x - 2$.
Substituting $\cos^2 x = \sin x$,we get:
$E = (\sin x)^6 + 3(\sin x)^5 + 3(\sin x)^4 + (\sin x)^3 - 2$.
This can be rewritten as:
$E = (\sin^2 x)^3 + 3(\sin^2 x)^2(\sin x) + 3(\sin^2 x)(\sin x)^2 + (\sin x)^3 - 2$.
Using the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,where $a = \sin^2 x$ and $b = \sin x$:
$E = (\sin^2 x + \sin x)^3 - 2$.
Since $\sin^2 x + \sin x = 1$,we have:
$E = (1)^3 - 2 = 1 - 2 = -1$.

(D) Given that $\sin x + \sin^2 x = 1.$
This implies $\sin x = 1 - \sin^2 x = \cos^2 x.$
Now,we need to evaluate the expression $\cos^8 x + 2\cos^6 x + \cos^4 x.$
Substituting $\cos^2 x = \sin x$ into the expression:
$= (\cos^2 x)^4 + 2(\cos^2 x)^3 + (\cos^2 x)^2$
$= (\sin x)^4 + 2(\sin x)^3 + (\sin x)^2$
$= \sin^4 x + 2\sin^3 x + \sin^2 x$
$= (\sin^2 x + \sin x)^2.$
Since $\sin x + \sin^2 x = 1,$ we have $(1)^2 = 1.$

(B) Given equations are:
$x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ ... $(i)$
$x \sin \alpha - y \cos \alpha = 0$ ... $(ii)$
From $(ii)$,we have $x \sin \alpha = y \cos \alpha$.
Substitute this into $(i)$:
$(x \sin \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$
$(y \cos \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$
$y \cos \alpha (\sin^2 \alpha + \cos^2 \alpha) = \sin \alpha \cos \alpha$
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we get:
$y \cos \alpha = \sin \alpha \cos \alpha$
Assuming $\cos \alpha \neq 0$,we get $y = \sin \alpha$.
Substituting $y = \sin \alpha$ into $x \sin \alpha = y \cos \alpha$,we get $x \sin \alpha = \sin \alpha \cos \alpha$,which implies $x = \cos \alpha$.
Therefore,$x^2 + y^2 = \cos^2 \alpha + \sin^2 \alpha = 1$.

(B) Let $x = (1 + \sin A)(1 + \sin B)(1 + \sin C)$ and $y = (1 - \sin A)(1 - \sin B)(1 - \sin C)$.
Given $x = y$.
Multiply both sides by $y$:
$xy = y^2$
$(1 + \sin A)(1 - \sin A)(1 + \sin B)(1 - \sin B)(1 + \sin C)(1 - \sin C) = y^2$
$(1 - \sin^2 A)(1 - \sin^2 B)(1 - \sin^2 C) = y^2$
$\cos^2 A \cos^2 B \cos^2 C = y^2$
Taking the square root on both sides:
$y = \pm \cos A \cos B \cos C$
Since $x = y$,each side is equal to $\pm \cos A \cos B \cos C$.

(A) Given: $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma ) = \tan \alpha \tan \beta \tan \gamma$ ... $(i)$
Let $x = (\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma )$ ... $(ii)$
Multiplying equation $(i)$ and $(ii)$,we get:
$(\sec^2 \alpha - \tan^2 \alpha )(\sec^2 \beta - \tan^2 \beta )(\sec^2 \gamma - \tan^2 \gamma ) = x \cdot (\tan \alpha \tan \beta \tan \gamma )$
Since $\sec^2 \theta - \tan^2 \theta = 1$,the left side becomes $1 \cdot 1 \cdot 1 = 1$.
So,$1 = x \cdot (\tan \alpha \tan \beta \tan \gamma )$
$x = \frac{1}{\tan \alpha \tan \beta \tan \gamma } = \cot \alpha \cot \beta \cot \gamma $

(D) Given that $\tan \theta - \cot \theta = a$ $(i)$ and $\sin \theta + \cos \theta = b$ $(ii)$.
Now,consider the expression $({b^2} - 1)^2({a^2} + 4)$.
Since $b = \sin \theta + \cos \theta$,we have $b^2 = (\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta = 1 + \sin 2\theta$.
Thus,$b^2 - 1 = \sin 2\theta$,so $(b^2 - 1)^2 = \sin^2 2\theta$.
Since $a = \tan \theta - \cot \theta$,we have $a^2 = (\tan \theta - \cot \theta)^2 = \tan^2 \theta + \cot^2 \theta - 2$.
Thus,$a^2 + 4 = \tan^2 \theta + \cot^2 \theta + 2 = (\tan \theta + \cot \theta)^2$.
Substituting these into the expression:
$(b^2 - 1)^2(a^2 + 4) = \sin^2 2\theta (\tan \theta + \cot \theta)^2$
$= (2 \sin \theta \cos \theta)^2 \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right)^2$
$= 4 \sin^2 \theta \cos^2 \theta \left( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right)^2$
$= 4 \sin^2 \theta \cos^2 \theta \left( \frac{1}{\sin \theta \cos \theta} \right)^2$
$= 4 \sin^2 \theta \cos^2 \theta \cdot \frac{1}{\sin^2 \theta \cos^2 \theta} = 4$.

(C) Given: $\tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2\tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1$.
Let $x = \tan^2 \alpha$,$y = \tan^2 \beta$,and $z = \tan^2 \gamma$.
Then the given equation is $xy + yz + zx + 2xyz = 1$.
We need to find $S = \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Using $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$,we have:
$S = \frac{x}{1+x} + \frac{y}{1+y} + \frac{z}{1+z}$.
$S = \frac{x(1+y)(1+z) + y(1+x)(1+z) + z(1+x)(1+y)}{(1+x)(1+y)(1+z)}$.
$S = \frac{x(1+y+z+yz) + y(1+x+z+xz) + z(1+x+y+xy)}{(1+x)(1+y)(1+z)}$.
$S = \frac{x+xy+xz+xyz + y+xy+yz+xyz + z+xz+yz+xyz}{(1+x)(1+y)(1+z)}$.
$S = \frac{(x+y+z) + 2(xy+yz+zx) + 3xyz}{(1+x)(1+y)(1+z)}$.
Since $xy+yz+zx+2xyz = 1$,we can substitute $xy+yz+zx = 1-2xyz$.
$S = \frac{(x+y+z) + 2(1-2xyz) + 3xyz}{1 + (x+y+z) + (xy+yz+zx) + xyz}$.
$S = \frac{x+y+z + 2 - 4xyz + 3xyz}{1 + (x+y+z) + (1-2xyz) + xyz} = \frac{x+y+z + 2 - xyz}{x+y+z + 2 - xyz} = 1$.

(B) Let $S = \sum_{k=1}^{36} \sin(k \times 10^\circ)$.
We know that $\sin(180^\circ + \theta) = -\sin \theta$.
Thus,$\sin(180^\circ + 10^\circ) = \sin 190^\circ = -\sin 10^\circ$.
Similarly,$\sin(180^\circ + 20^\circ) = \sin 200^\circ = -\sin 20^\circ$,and so on up to $\sin(180^\circ + 170^\circ) = \sin 350^\circ = -\sin 170^\circ$.
Also,$\sin 180^\circ = 0$ and $\sin 360^\circ = 0$.
The sum can be written as $(\sin 10^\circ + \sin 190^\circ) + (\sin 20^\circ + \sin 200^\circ) + \dots + (\sin 170^\circ + \sin 350^\circ) + \sin 180^\circ + \sin 360^\circ$.
Since each pair sums to $0$ and $\sin 180^\circ = \sin 360^\circ = 0$,the total sum is $0$.

(A) Given $\alpha = 22^\circ 30' = \frac{\pi}{8}$.
Let $P = (1 + \cos \alpha)(1 + \cos 3\alpha)(1 + \cos 5\alpha)(1 + \cos 7\alpha)$.
Note that $3\alpha = 67^\circ 30'$,$5\alpha = 112^\circ 30'$,and $7\alpha = 157^\circ 30'$.
We know $\cos 5\alpha = \cos(180^\circ - 67^\circ 30') = -\cos 3\alpha$ and $\cos 7\alpha = \cos(180^\circ - 22^\circ 30') = -\cos \alpha$.
So,$P = (1 + \cos \alpha)(1 - \cos \alpha)(1 + \cos 3\alpha)(1 - \cos 3\alpha)$.
$P = (1 - \cos^2 \alpha)(1 - \cos^2 3\alpha) = \sin^2 \alpha \sin^2 3\alpha$.
Since $\alpha = 22.5^\circ$,$\sin^2 \alpha = \frac{1 - \cos 45^\circ}{2} = \frac{1 - 1/\sqrt{2}}{2} = \frac{\sqrt{2} - 1}{2\sqrt{2}}$.
Since $3\alpha = 67.5^\circ$,$\sin^2 3\alpha = \frac{1 - \cos 135^\circ}{2} = \frac{1 - (-1/\sqrt{2})}{2} = \frac{\sqrt{2} + 1}{2\sqrt{2}}$.
$P = \left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right) \left(\frac{\sqrt{2} + 1}{2\sqrt{2}}\right) = \frac{2 - 1}{8} = \frac{1}{8}$.

(C) We use the identities $\sin^2 \theta + \cos^2 \theta = 1$,$\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$,and $\sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta$.
Substituting these into the expression:
$6(1 - 3\sin^2 \theta \cos^2 \theta) - 9(1 - 2\sin^2 \theta \cos^2 \theta) + 4$
$= 6 - 18\sin^2 \theta \cos^2 \theta - 9 + 18\sin^2 \theta \cos^2 \theta + 4$
$= 6 - 9 + 4 = 1$.

(A) Given $A = 130^\circ$.
$x = \sin 130^\circ + \cos 130^\circ$.
Using the identity $\sin \theta = \cos(90^\circ - \theta)$,we have $\sin 130^\circ = \cos(90^\circ - 130^\circ) = \cos(-40^\circ) = \cos 40^\circ$.
So,$x = \cos 40^\circ + \cos 130^\circ$.
Using the sum-to-product formula $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$x = 2 \cos(\frac{40^\circ + 130^\circ}{2}) \cos(\frac{40^\circ - 130^\circ}{2}) = 2 \cos(85^\circ) \cos(-45^\circ) = 2 \cos 85^\circ \cos 45^\circ$.
Since $85^\circ$ is in the first quadrant,$\cos 85^\circ > 0$.
Since $45^\circ$ is in the first quadrant,$\cos 45^\circ > 0$.
Therefore,$x = 2 \cos 85^\circ \cos 45^\circ > 0$.

(B) Given expression: $E = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}}$
Taking the common denominator: $E = \frac{(1 - \cos \alpha) + (1 + \cos \alpha)}{\sqrt{(1 + \cos \alpha)(1 - \cos \alpha)}}$
Simplifying the numerator and denominator: $E = \frac{2}{\sqrt{1 - \cos^2 \alpha}} = \frac{2}{\sqrt{\sin^2 \alpha}} = \frac{2}{|\sin \alpha|}$
Since $\pi < \alpha < \frac{3\pi}{2}$,$\alpha$ lies in the third quadrant,where $\sin \alpha$ is negative.
Therefore,$|\sin \alpha| = -\sin \alpha$.
Thus,$E = \frac{2}{-\sin \alpha} = -\frac{2}{\sin \alpha}$.

(A) Let the two parts be $A$ and $B$ such that $A + B = \theta$ and $A - B = \phi$.
Given that $\tan A = k \tan B$,we have $\frac{\tan A}{\tan B} = k$.
Expanding the tangents in terms of sine and cosine:
$\frac{\sin A \cos B}{\cos A \sin B} = \frac{k}{1}$.
Applying the componendo and dividendo rule:
$\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{k + 1}{k - 1}$.
Using the identity $\frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B} = \frac{\sin(A + B)}{\sin(A - B)}$,we get:
$\frac{\sin(A + B)}{\sin(A - B)} = \frac{k + 1}{k - 1}$.
Substituting $A + B = \theta$ and $A - B = \phi$:
$\frac{\sin \theta}{\sin \phi} = \frac{k + 1}{k - 1}$.
Therefore,$\sin \theta = \frac{k + 1}{k - 1} \sin \phi$.

(A) Given $\pi < \alpha < \frac{3\pi}{2}$,$\alpha$ lies in the third quadrant,so $\sin \alpha < 0$.
Let $E = \sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$.
Using $\sin 2\alpha = 2\sin \alpha \cos \alpha$,we have $\sin^2 2\alpha = 4\sin^2 \alpha \cos^2 \alpha$.
$E = \sqrt{4\sin^4 \alpha + 4\sin^2 \alpha \cos^2 \alpha} + 2(1 + \cos(\frac{\pi}{2} - \alpha))$
$E = \sqrt{4\sin^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} + 2(1 + \sin \alpha)$
$E = \sqrt{4\sin^2 \alpha} + 2 + 2\sin \alpha$
$E = 2|\sin \alpha| + 2 + 2\sin \alpha$
Since $\alpha$ is in the third quadrant,$\sin \alpha < 0$,so $|\sin \alpha| = -\sin \alpha$.
$E = 2(-\sin \alpha) + 2 + 2\sin \alpha = 2$.

(C) Let the expression be $E = (\sec A + \tan A - 1)(\sec A - (\tan A - 1)) - 2\tan A$.
Using the identity $(x+y)(x-y) = x^2 - y^2$,we rewrite the first part:
$E = (\sec A + (\tan A - 1))(\sec A - (\tan A - 1)) - 2\tan A$
$E = \sec^2 A - (\tan A - 1)^2 - 2\tan A$
$E = \sec^2 A - (\tan^2 A - 2\tan A + 1) - 2\tan A$
$E = \sec^2 A - \tan^2 A + 2\tan A - 1 - 2\tan A$
Since $\sec^2 A - \tan^2 A = 1$,we have:
$E = 1 - 1 = 0$.

(C) Given $\tan A = 2\tan B + \cot B.$
We know that $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.$
Substituting the value of $\tan A$:
$2\tan (A - B) = 2 \left( \frac{2\tan B + \cot B - \tan B}{1 + (2\tan B + \cot B)\tan B} \right)$
$= 2 \left( \frac{\tan B + \cot B}{1 + 2\tan^2 B + \cot B \tan B} \right)$
Since $\cot B \tan B = 1$:
$= 2 \left( \frac{\tan B + \cot B}{1 + 2\tan^2 B + 1} \right) = 2 \left( \frac{\tan B + \cot B}{2 + 2\tan^2 B} \right)$
$= 2 \left( \frac{\tan B + \cot B}{2(1 + \tan^2 B)} \right) = \frac{\tan B + \cot B}{\sec^2 B}$
$= \frac{\tan B}{\sec^2 B} + \frac{\cot B}{\sec^2 B} = \sin B \cos B + \frac{\cos B}{\sin B} \cos^2 B = \sin B \cos B + \frac{\cos^3 B}{\sin B}$
Alternatively,simplifying the expression directly:
$= \frac{\tan B + \frac{1}{\tan B}}{1 + \tan^2 B} = \frac{\frac{\tan^2 B + 1}{\tan B}}{1 + \tan^2 B} = \frac{1}{\tan B} = \cot B.$

(A) Given that $\sin A = \sin B$ and $\cos A = \cos B.$
Subtracting the squares of the equations:
$\sin^2 A + \cos^2 A = \sin^2 B + \cos^2 B$
$1 = 1$ (This is always true).
Consider the difference formula:
$\cos A = \cos B \Rightarrow \cos A - \cos B = 0$
$-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} = 0$
Also,$\sin A = \sin B \Rightarrow \sin A - \sin B = 0$
$2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} = 0$
From these,we can see that $\sin \frac{A-B}{2} = 0$ must hold for both conditions to be satisfied simultaneously.

(C) Given $A - B = \frac{\pi}{4}$.
Taking tangent on both sides,$\tan(A - B) = \tan\frac{\pi}{4} = 1$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = 1$.
This implies $\tan A - \tan B = 1 + \tan A \tan B$.
Rearranging,$\tan A - \tan B - \tan A \tan B = 1$.
Adding $1$ to both sides,$\tan A - \tan B - \tan A \tan B + 1 = 1 + 1 = 2$.
Factoring the expression,$(1 + \tan A)(1 - \tan B) = 2$.
Since $y = (1 + \tan A)(1 - \tan B)$,we have $y = 2$.
Therefore,$(y + 1)^{y + 1} = (2 + 1)^{2 + 1} = 3^3 = 27$.

(D) Let $X = \frac{\cot A}{1 + \cot A} \cdot \frac{\cot B}{1 + \cot B}$.
Converting to tangent form: $X = \frac{1/\tan A}{1 + 1/\tan A} \cdot \frac{1/\tan B}{1 + 1/\tan B} = \frac{1}{\tan A + 1} \cdot \frac{1}{\tan B + 1}$.
$X = \frac{1}{\tan A \tan B + \tan A + \tan B + 1}$.
Given $A + B = 225^\circ$,we have $\tan(A + B) = \tan(225^\circ) = 1$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$,we get $\tan A + \tan B = 1 - \tan A \tan B$.
Substituting this into the expression for $X$:
$X = \frac{1}{\tan A \tan B + (1 - \tan A \tan B) + 1} = \frac{1}{1 + 1} = \frac{1}{2}$.

(D) Given expression: $\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$
$= \frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{2(\frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
Using $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
$= \frac{2(\sin 30^\circ \cos 10^\circ - \cos 30^\circ \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{2 \sin(30^\circ - 10^\circ)}{\sin 10^\circ \cos 10^\circ}$
$= \frac{2 \sin 20^\circ}{\sin 10^\circ \cos 10^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{4 \sin 20^\circ}{2 \sin 10^\circ \cos 10^\circ}$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{4 \sin 20^\circ}{\sin 20^\circ} = 4$.

(B) Given $\sin (\theta + \alpha ) = a$ and $\sin (\theta + \beta ) = b$.
Let $x = \theta + \alpha$ and $y = \theta + \beta$. Then $\alpha - \beta = x - y$.
We have $\sin x = a$ and $\sin y = b$. Thus $\cos x = \pm \sqrt{1-a^2}$ and $\cos y = \pm \sqrt{1-b^2}$.
$\cos (\alpha - \beta ) = \cos (x - y) = \cos x \cos y + \sin x \sin y = \pm \sqrt{1-a^2}\sqrt{1-b^2} + ab$.
Let $C = \cos (\alpha - \beta )$. Then $C - ab = \pm \sqrt{1-a^2}\sqrt{1-b^2}$.
Squaring both sides: $(C - ab)^2 = (1-a^2)(1-b^2) = 1 - a^2 - b^2 + a^2b^2$.
$C^2 - 2abC + a^2b^2 = 1 - a^2 - b^2 + a^2b^2$.
$C^2 - 2abC = 1 - a^2 - b^2$.
We need to evaluate $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta ) = 2\cos^2 (\alpha - \beta ) - 1 - 4ab\cos (\alpha - \beta )$.
$= 2(C^2 - 2abC) - 1$.
Substituting $C^2 - 2abC = 1 - a^2 - b^2$:
$= 2(1 - a^2 - b^2) - 1 = 2 - 2a^2 - 2b^2 - 1 = 1 - 2a^2 - 2b^2$.

(C) Let the expression be $E = \cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$.
Using the identity $2\cos A\cos B = \cos(A - B) + \cos(A + B)$,we have:
$E = \cos^2(A - B) + \cos^2 B - \cos(A - B)[\cos(A - B) + \cos(A + B)]$
$E = \cos^2(A - B) + \cos^2 B - \cos^2(A - B) - \cos(A - B)\cos(A + B)$
$E = \cos^2 B - \cos(A - B)\cos(A + B)$
Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B$:
$E = \cos^2 B - (\cos^2 A - \sin^2 B)$
$E = \cos^2 B + \sin^2 B - \cos^2 A$
Since $\cos^2 B + \sin^2 B = 1$,we get:
$E = 1 - \cos^2 A = \sin^2 A$
Since the result $\sin^2 A$ depends only on $A$,the expression is dependent on $A$.