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

(A) Given $\sin A = n \sin (A + 2B)$.
We can write this as $\frac{\sin A}{\sin (A + 2B)} = n$.
Applying componendo and dividendo:
$\frac{\sin (A + 2B) + \sin A}{\sin (A + 2B) - \sin A} = \frac{1 + n}{1 - n}$.
Using the sum-to-product formulas:
$\frac{2 \sin(A + B) \cos B}{2 \cos(A + B) \sin B} = \frac{1 + n}{1 - n}$.
$\tan (A + B) \cot B = \frac{1 + n}{1 - n}$.
Therefore,$\tan (A + B) = \frac{1 + n}{1 - n} \tan B$.

(A) Given: $\sin A + \sin B = x$ $(1)$ and $\cos A + \cos B = y$ $(2)$.
Using sum-to-product formulas:
$2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2}) = x$ $(3)$
$2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2}) = y$ $(4)$
Dividing $(3)$ by $(4)$:
$\tan(\frac{A+B}{2}) = \frac{x}{y}$.
We know that $\sin(A+B) = \frac{2 \tan(\frac{A+B}{2})}{1 + \tan^2(\frac{A+B}{2})}$.
Substituting the value:
$\sin(A+B) = \frac{2(x/y)}{1 + (x/y)^2} = \frac{2x/y}{(y^2 + x^2)/y^2} = \frac{2xy}{x^2 + y^2}$.

(A) Given $\sin \theta = \frac{1}{2} (x + \frac{1}{x})$.
Since the range of $\sin \theta$ is $[-1, 1]$,and for any real $x \neq 0$,$|x + \frac{1}{x}| \geq 2$,we have $|\sin \theta| = \frac{1}{2} |x + \frac{1}{x}| \geq 1$.
Thus,$\sin \theta$ can only be $1$ or $-1$.
If $\sin \theta = 1$,then $x + \frac{1}{x} = 2$,which implies $x = 1$.
Then $\sin 3 \theta = \sin(3 \times 90^\circ) = \sin 270^\circ = -1$.
Also,$\frac{1}{2} (x^3 + \frac{1}{x^3}) = \frac{1}{2} (1^3 + \frac{1}{1^3}) = \frac{1}{2} (2) = 1$.
Therefore,$\sin 3 \theta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = -1 + 1 = 0$.
If $\sin \theta = -1$,then $x + \frac{1}{x} = -2$,which implies $x = -1$.
Then $\sin 3 \theta = \sin(3 \times 270^\circ) = \sin 810^\circ = \sin 90^\circ = 1$.
Also,$\frac{1}{2} (x^3 + \frac{1}{x^3}) = \frac{1}{2} ((-1)^3 + \frac{1}{(-1)^3}) = \frac{1}{2} (-2) = -1$.
Therefore,$\sin 3 \theta + \frac{1}{2} (x^3 + \frac{1}{x^3}) = 1 - 1 = 0$.

(A) Given $\sin(\alpha+\beta)=1$. Since $\alpha, \beta \in [0, \frac{\pi}{2}]$,$\alpha+\beta = \frac{\pi}{2}$.
Given $\sin(\alpha-\beta)=\frac{1}{2}$. Since $\alpha, \beta \in [0, \frac{\pi}{2}]$,$\alpha-\beta = \frac{\pi}{6}$.
Adding the two equations: $2\alpha = \frac{\pi}{2} + \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \implies \alpha = \frac{\pi}{3}$.
Subtracting the two equations: $2\beta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \implies \beta = \frac{\pi}{6}$.
Now,calculate $\tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta)$:
$\alpha+2\beta = \frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.
$2\alpha+\beta = 2(\frac{\pi}{3}) + \frac{\pi}{6} = \frac{2\pi}{3} + \frac{\pi}{6} = \frac{5\pi}{6}$.
$\tan(\frac{2\pi}{3}) = -\sqrt{3}$ and $\tan(\frac{5\pi}{6}) = -\frac{1}{\sqrt{3}}$.
Product $= (-\sqrt{3}) \cdot (-\frac{1}{\sqrt{3}}) = 1$.

(C) Given $\tan \theta = k \tan \phi$ and $\theta + \phi = \alpha$.
$\frac{\tan \theta}{\tan \phi} = \frac{k}{1}$
Applying Componendo and Dividendo:
$\frac{\tan \theta + \tan \phi}{\tan \theta - \tan \phi} = \frac{k+1}{k-1}$
Converting to sine and cosine:
$\frac{\frac{\sin \theta}{\cos \theta} + \frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta} - \frac{\sin \phi}{\cos \phi}} = \frac{k+1}{k-1}$
$\frac{\sin \theta \cos \phi + \cos \theta \sin \phi}{\sin \theta \cos \phi - \cos \theta \sin \phi} = \frac{k+1}{k-1}$
Using the identity $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:
$\frac{\sin(\theta + \phi)}{\sin(\theta - \phi)} = \frac{k+1}{k-1}$
Substituting $\theta + \phi = \alpha$:
$\frac{\sin \alpha}{\sin(\theta - \phi)} = \frac{k+1}{k-1}$
Therefore,$\sin(\theta - \phi) = \left(\frac{k-1}{k+1}\right) \sin \alpha$.

(B) Given equations are:
$\cos x + \cos y = -\cos \alpha$ ... $(1)$
$\sin x + \sin y = -\sin \alpha$ ... $(2)$
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\cos \alpha$ ... $(3)$
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\sin \alpha$ ... $(4)$
Dividing equation $(4)$ by equation $(3)$:
$\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{-\sin \alpha}{-\cos \alpha}$
$\tan \left(\frac{x+y}{2}\right) = \tan \alpha$
Therefore,$\cot \left(\frac{x+y}{2}\right) = \cot \alpha$.

(A) Given equation: $(1+\sqrt{1+x}) \tan x = 1+\sqrt{1-x}$
$\tan x = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$
Let $x = \sin \theta$.
$\tan x = \frac{1+\sqrt{1-\sin \theta}}{1+\sqrt{1+\sin \theta}} = \frac{1+\sqrt{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2}}{1+\sqrt{(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2}}$
$= \frac{1+\cos \frac{\theta}{2} - \sin \frac{\theta}{2}}{1+\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$
$= \frac{2\cos^2 \frac{\theta}{4} - 2\sin \frac{\theta}{4}\cos \frac{\theta}{4}}{2\cos^2 \frac{\theta}{4} + 2\sin \frac{\theta}{4}\cos \frac{\theta}{4}}$
$= \frac{2\cos \frac{\theta}{4}(\cos \frac{\theta}{4} - \sin \frac{\theta}{4})}{2\cos \frac{\theta}{4}(\cos \frac{\theta}{4} + \sin \frac{\theta}{4})} = \frac{1 - \tan \frac{\theta}{4}}{1 + \tan \frac{\theta}{4}} = \tan(\frac{\pi}{4} - \frac{\theta}{4})$
Thus,$x = \frac{\pi}{4} - \frac{\theta}{4} \Rightarrow 4x = \pi - \theta$.
$\sin 4x = \sin(\pi - \theta) = \sin \theta = x$.

(B) Given expression: $E = \frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}} + \tan 147^{\circ}$
Divide the numerator and denominator of the first term by $\cos 12^{\circ}$:
$E = \frac{1-\tan 12^{\circ}}{1+\tan 12^{\circ}} + \tan(180^{\circ}-33^{\circ})$
$E = \tan(45^{\circ}-12^{\circ}) - \tan 33^{\circ}$
$E = \tan 33^{\circ} - \tan 33^{\circ} = 0$

(C) Given that $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A$.$P$.
$\therefore 2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$
Using the formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:
$2 \sin (z+x-y) = 2 \sin \left( \frac{y+z-x+x+y-z}{2} \right) \cos \left( \frac{y+z-x-x-y+z}{2} \right)$
$2 \sin (z+x-y) = 2 \sin y \cos (z-x)$
Dividing both sides by $\cos (y+z-x) \cos (z+x-y) \cos (x+y-z)$ is not direct,so we use the property $\frac{\sin (A-B)}{\cos A \cos B} = \tan A - \tan B$.
By simplifying the $A$.$P$. condition,we get $\tan x, \tan y, \tan z$ are in $A$.$P$.
Therefore,$2 \tan y = \tan x + \tan z$.

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$
$= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$ where $A=60^{\circ}$ and $B=20^{\circ}$:
$= \frac{2(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2}(2 \sin 20^{\circ} \cos 20^{\circ})}$
$= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

(A) We have the expression: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$
Since $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$ and $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$.
Substituting these,the expression becomes: $\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)$
$= \left(1-\cos^2 \frac{\pi}{8}\right)\left(1-\cos^2 \frac{3 \pi}{8}\right)$
$= \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$
$= \frac{1}{4} \left(2 \sin \frac{\pi}{8} \sin \frac{3 \pi}{8}\right)^2$
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$= \frac{1}{4} \left(\cos \frac{2\pi}{8} - \cos \frac{4\pi}{8}\right)^2$
$= \frac{1}{4} \left(\cos \frac{\pi}{4} - \cos \frac{\pi}{2}\right)^2$
$= \frac{1}{4} \left(\frac{1}{\sqrt{2}} - 0\right)^2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

(A) Given $\cos 2B = \frac{\cos(A+C)}{\cos(A-C)}$.
Using the formula $\cos 2B = \frac{1-\tan^2 B}{1+\tan^2 B}$ and expanding the right side:
$\frac{1-\tan^2 B}{1+\tan^2 B} = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$.
Dividing numerator and denominator by $\cos A \cos C$:
$\frac{1-\tan^2 B}{1+\tan^2 B} = \frac{1-\tan A \tan C}{1+\tan A \tan C}$.
By componendo and dividendo or cross-multiplication:
$(1-\tan^2 B)(1+\tan A \tan C) = (1+\tan^2 B)(1-\tan A \tan C)$.
$1 + \tan A \tan C - \tan^2 B - \tan^2 B \tan A \tan C = 1 - \tan A \tan C + \tan^2 B - \tan^2 B \tan A \tan C$.
Simplifying the equation:
$2 \tan A \tan C = 2 \tan^2 B$.
$\tan^2 B = \tan A \tan C$.
Therefore,$\tan A, \tan B, \tan C$ are in Geometric Progression $(G.P.)$.

(A) Given,$\tan A + \cot A = 2$.
Squaring both sides,we get:
$(\tan A + \cot A)^{2} = 2^{2}$
$\tan^{2} A + \cot^{2} A + 2 \tan A \cot A = 4$
Since $\tan A \cot A = 1$,we have:
$\tan^{2} A + \cot^{2} A + 2 = 4$
$\tan^{2} A + \cot^{2} A = 2$
Now,squaring both sides again:
$(\tan^{2} A + \cot^{2} A)^{2} = 2^{2}$
$\tan^{4} A + \cot^{4} A + 2 \tan^{2} A \cot^{2} A = 4$
$\tan^{4} A + \cot^{4} A + 2(1)^{2} = 4$
$\tan^{4} A + \cot^{4} A = 4 - 2 = 2$.

(B) Given that $\cos x + \cos^2 x = 1$.
From this,we have $\cos x = 1 - \cos^2 x$.
Using the identity $\sin^2 x + \cos^2 x = 1$,we get $\sin^2 x = \cos x$.
Now,we need to find the value of $\sin^2 x + \sin^4 x$.
Substituting $\sin^2 x = \cos x$,we get $\sin^2 x + \sin^4 x = \cos x + (\cos x)^2$.
Since $\cos x + \cos^2 x = 1$,the value is $1$.

(C) Let $S = \log _{10} \tan 1^{\circ} + \log _{10} \tan 2^{\circ} + \ldots + \log _{10} \tan 89^{\circ}$.
Using the property $\log a + \log b = \log(ab)$,we have:
$S = \log _{10} (\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \cdot \ldots \cdot \tan 89^{\circ})$.
We know that $\tan \theta \cdot \tan(90^{\circ} - \theta) = \tan \theta \cdot \cot \theta = 1$.
Pairing the terms: $(\tan 1^{\circ} \cdot \tan 89^{\circ}) = 1, (\tan 2^{\circ} \cdot \tan 88^{\circ}) = 1, \ldots, (\tan 44^{\circ} \cdot \tan 46^{\circ}) = 1$.
The middle term is $\tan 45^{\circ} = 1$.
Thus,the product is $1 \cdot 1 \cdot \ldots \cdot 1 = 1$.
Therefore,$S = \log _{10} (1) = 0$.
The expression becomes $e^S = e^0 = 1$.

(B) Let the expression be $E = \cot 12^{\circ} \cot 102^{\circ} + \cot 102^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$.
We know that $\cot 102^{\circ} = \cot(90^{\circ} + 12^{\circ}) = -\tan 12^{\circ}$.
Substituting this into the expression:
$E = \cot 12^{\circ} (-\tan 12^{\circ}) + (-\tan 12^{\circ}) \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$
$E = -1 - \tan 12^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$
$E = -1 + \cot 66^{\circ} (\cot 12^{\circ} - \tan 12^{\circ})$
Using the identity $\cot \theta - \tan \theta = 2 \cot 2\theta$:
$E = -1 + \cot 66^{\circ} (2 \cot 24^{\circ})$
Since $\cot 24^{\circ} = \cot(90^{\circ} - 66^{\circ}) = \tan 66^{\circ}$:
$E = -1 + 2 \cot 66^{\circ} \tan 66^{\circ}$
$E = -1 + 2(1) = 1$.

(B) Let $f(x) = 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin ^{6} x+\cos ^{6} x)$.
First,simplify each term:
$3(\sin x-\cos x)^{4} = 3(1-2\sin x\cos x)^{2} = 3(1+4\sin^{2}x\cos^{2}x-4\sin x\cos x) = 3+12\sin^{2}x\cos^{2}x-12\sin x\cos x$.
$6(\sin x+\cos x)^{2} = 6(1+2\sin x\cos x) = 6+12\sin x\cos x$.
$4(\sin^{6}x+\cos^{6}x) = 4(\sin^{2}x+\cos^{2}x)(\sin^{4}x+\cos^{4}x-\sin^{2}x\cos^{2}x) = 4(1)((\sin^{2}x+\cos^{2}x)^{2}-3\sin^{2}x\cos^{2}x) = 4(1-3\sin^{2}x\cos^{2}x) = 4-12\sin^{2}x\cos^{2}x$.
Summing these up:
$f(x) = (3+12\sin^{2}x\cos^{2}x-12\sin x\cos x) + (6+12\sin x\cos x) + (4-12\sin^{2}x\cos^{2}x)$.
$f(x) = 3+6+4 + (12\sin^{2}x\cos^{2}x-12\sin^{2}x\cos^{2}x) + (-12\sin x\cos x+12\sin x\cos x) = 13$.

(D) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$
$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} \right)}{\frac{1}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$

(A) The given expression is $S = \sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$.
We know that $\sin(360^{\circ} - \theta) = -\sin \theta$.
Thus,$\sin 359^{\circ} = \sin(360^{\circ} - 1^{\circ}) = -\sin 1^{\circ}$.
Similarly,$\sin 358^{\circ} = -\sin 2^{\circ}$,and so on.
We can pair the terms as $(\sin 1^{\circ} + \sin 359^{\circ}) + (\sin 2^{\circ} + \sin 358^{\circ}) + \ldots + (\sin 179^{\circ} + \sin 181^{\circ}) + \sin 180^{\circ}$.
Since $\sin(180^{\circ} + \theta) = -\sin \theta$,we have $\sin 181^{\circ} = -\sin 1^{\circ}$,$\sin 182^{\circ} = -\sin 2^{\circ}$,etc.
Each pair sums to $0$,and $\sin 180^{\circ} = 0$.
Therefore,the total sum is $0$.

(B) Given $a = \frac{x}{y-z}$,$b = \frac{y}{z-x}$,$c = \frac{z}{x-y}$.
Let us test with values $x=1, y=2, z=4$ (ensuring denominators are non-zero).
$a = \frac{1}{2-4} = -\frac{1}{2}$
$b = \frac{2}{4-1} = \frac{2}{3}$
$c = \frac{4}{1-2} = -4$
Now calculate $ab + bc + ca + abc$:
$ab = (-\frac{1}{2})(\frac{2}{3}) = -\frac{1}{3}$
$bc = (\frac{2}{3})(-4) = -\frac{8}{3}$
$ca = (-4)(-\frac{1}{2}) = 2$
$abc = (-\frac{1}{2})(\frac{2}{3})(-4) = \frac{4}{3}$
Sum $= -\frac{1}{3} - \frac{8}{3} + 2 + \frac{4}{3} = \frac{-1-8+4}{3} + 2 = -\frac{5}{3} + 2 = \frac{1}{3}$.
Note: The original expression $ab+bc+ca$ evaluates to $-1$ for these specific variables.

(D) Given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$.
By the relation between roots and coefficients,we have:
$\tan A + \tan B = p$
$\tan A \tan B = q$
Using the trigonometric identity for $\tan(A+B)$:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{p}{1-q}$
Now,we know that $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$.
Substituting $\tan(A+B) = \frac{p}{1-q}$:
$\sin^2(A+B) = \frac{(\frac{p}{1-q})^2}{1 + (\frac{p}{1-q})^2} = \frac{\frac{p^2}{(1-q)^2}}{\frac{(1-q)^2 + p^2}{(1-q)^2}} = \frac{p^2}{p^2+(1-q)^2}$

(C) Given equations are:
$y^2+z^2=3yz \Rightarrow \frac{y}{z}+\frac{z}{y}=3$ $(i)$
$z^2+x^2=8zx \Rightarrow \frac{z}{x}+\frac{x}{z}=8$ $(ii)$
$x^2+y^2=4xy \Rightarrow \frac{x}{y}+\frac{y}{x}=4$ $(iii)$
Multiplying $(i)$ and $(iii)$:
$\left(\frac{y}{z}+\frac{z}{y}\right)\left(\frac{x}{y}+\frac{y}{x}\right) = \frac{xy}{zy} + \frac{y^2}{xz} + \frac{z}{y} \cdot \frac{x}{y} + \frac{z}{y} \cdot \frac{y}{x} = \frac{x}{z} + \frac{y^2}{xz} + \frac{xz}{y^2} + \frac{z}{x} = 12$
$\left(\frac{x}{z}+\frac{z}{x}\right) + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$
Substituting the value from $(ii)$:
$8 + \left(\frac{y^2}{xz}+\frac{xz}{y^2}\right) = 12$
$\frac{y^2}{xz}+\frac{xz}{y^2} = 12 - 8 = 4$

(C) Given equations are:
$y^2+z^2=a y z \Rightarrow \frac{y}{z}+\frac{z}{y}=a$ $(i)$
$z^2+x^2=b z x \Rightarrow \frac{z}{x}+\frac{x}{z}=b$ $(ii)$
$x^2+y^2=c x y \Rightarrow \frac{x}{y}+\frac{y}{x}=c$ $(iii)$
Multiplying $(i)$ and $(iii)$:
$(\frac{y}{z}+\frac{z}{y})(\frac{x}{y}+\frac{y}{x}) = ac$
$\frac{x}{z} + \frac{y^2}{zx} + \frac{zx}{y^2} + \frac{z}{x} = ac$
From $(ii)$,we know $\frac{z}{x}+\frac{x}{z}=b$. Substituting this into the expression:
$b + \frac{y^2}{zx} + \frac{zx}{y^2} = ac$
$\frac{y^2}{zx} + \frac{zx}{y^2} = ac-b$

(B) Let $P = \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \ldots \cdot \tan 89^{\circ}$.
Using the property $\tan(90^{\circ} - \theta) = \cot \theta$,we can write the product as:
$P = (\tan 1^{\circ} \cdot \tan 89^{\circ}) \cdot (\tan 2^{\circ} \cdot \tan 88^{\circ}) \cdot \ldots \cdot (\tan 44^{\circ} \cdot \tan 46^{\circ}) \cdot \tan 45^{\circ}$.
Since $\tan \theta \cdot \tan(90^{\circ} - \theta) = \tan \theta \cdot \cot \theta = 1$,each pair equals $1$.
Thus,$P = 1 \cdot 1 \cdot \ldots \cdot 1 \cdot \tan 45^{\circ} = 1 \cdot 1 = 1$.
The geometric mean of $n$ terms $a_1, a_2, \ldots, a_n$ is $(a_1 \cdot a_2 \cdot \ldots \cdot a_n)^{1/n}$.
Here,$n = 89$,so the geometric mean is $(P)^{1/89} = (1)^{1/89} = 1$.
Hence,option $B$ is correct.

(D) We know that $\cot 18^{\circ} \cot 36^{\circ}+1 = \frac{\cos 18^{\circ}}{\sin 18^{\circ}} \cdot \frac{\cos 36^{\circ}}{\sin 36^{\circ}}+1$.
Using $\cos 36^{\circ} = 1-2\sin^2 18^{\circ}$,we get:
$\frac{\cos 18^{\circ}(1-2\sin^2 18^{\circ})}{\sin 18^{\circ} \cdot 2 \sin 18^{\circ} \cos 18^{\circ}}+1 = \frac{1-2\sin^2 18^{\circ}}{2\sin^2 18^{\circ}}+1 = \frac{1}{2\sin^2 18^{\circ}}$.
Since $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then $\sin^2 18^{\circ} = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.
Substituting this value:
$\frac{1}{2(\frac{3-\sqrt{5}}{8})} = \frac{4}{3-\sqrt{5}} = \frac{4(3+\sqrt{5})}{9-5} = \frac{4(3+\sqrt{5})}{4} = 3+\sqrt{5}$.

(C) We use the trigonometric identity: $\cos^3 x + \cos^3(120^{\circ} - x) + \cos^3(120^{\circ} + x) = \frac{3}{4} \cos(3x)$.
Given the expression $\cos^3 10^{\circ} + \cos^3 110^{\circ} + \cos^3 130^{\circ}$,we can rewrite it as:
$\cos^3 10^{\circ} + \cos^3(120^{\circ} - 10^{\circ}) + \cos^3(120^{\circ} + 10^{\circ})$.
Here,$x = 10^{\circ}$.
Applying the identity:
$= \frac{3}{4} \cos(3 \times 10^{\circ})$
$= \frac{3}{4} \cos 30^{\circ}$
$= \frac{3}{4} \times \frac{\sqrt{3}}{2}$
$= \frac{3\sqrt{3}}{8}$.

(A) We know that $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Consider the identity $\tan 60^{\circ} = \tan(40^{\circ} + 20^{\circ}) = \frac{\tan 40^{\circ} + \tan 20^{\circ}}{1 - \tan 40^{\circ} \tan 20^{\circ}} = \sqrt{3}$.
This implies $\tan 40^{\circ} + \tan 20^{\circ} = \sqrt{3}(1 - \tan 40^{\circ} \tan 20^{\circ}) = \sqrt{3} - \sqrt{3} \tan 40^{\circ} \tan 20^{\circ}$.
Rearranging gives $\tan 40^{\circ} + \tan 20^{\circ} + \sqrt{3} \tan 40^{\circ} \tan 20^{\circ} = \sqrt{3}$.
Now consider $\tan(56^{\circ} - 11^{\circ}) = \tan 45^{\circ} = 1$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have $\frac{\tan 56^{\circ} - \tan 11^{\circ}}{1 + \tan 56^{\circ} \tan 11^{\circ}} = 1$.
This implies $\tan 56^{\circ} - \tan 11^{\circ} = 1 + \tan 56^{\circ} \tan 11^{\circ}$,or $\tan 56^{\circ} - \tan 11^{\circ} - \tan 56^{\circ} \tan 11^{\circ} = 1$.
Substituting these into the original expression: $(\tan 40^{\circ} + \tan 20^{\circ} + \sqrt{3} \tan 40^{\circ} \tan 20^{\circ}) + (\tan 11^{\circ} - \tan 56^{\circ} + \tan 56^{\circ} \tan 11^{\circ})$.
From the second identity,$\tan 56^{\circ} - \tan 11^{\circ} - \tan 56^{\circ} \tan 11^{\circ} = 1$,so $\tan 11^{\circ} - \tan 56^{\circ} + \tan 56^{\circ} \tan 11^{\circ} = -1$.
Thus,the total value is $\sqrt{3} - 1$.

(C) Let $E = \cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have:
$E = \frac{1 + \cos 152^{\circ}}{2} + \frac{1 + \cos 32^{\circ}}{2} - \cos 76^{\circ} \cos 16^{\circ}$
$E = 1 + \frac{1}{2} (\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$\cos 152^{\circ} + \cos 32^{\circ} = 2 \cos 92^{\circ} \cos 60^{\circ} = 2 \cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$\cos 76^{\circ} \cos 16^{\circ} = \frac{1}{2} (\cos 92^{\circ} + \cos 60^{\circ}) = \frac{1}{2} \cos 92^{\circ} + \frac{1}{4}$
Substituting these back:
$E = 1 + \frac{1}{2} \cos 92^{\circ} - (\frac{1}{2} \cos 92^{\circ} + \frac{1}{4})$
$E = 1 - \frac{1}{4} = \frac{3}{4}$

(A) Given: $\sin \alpha = \frac{6}{5} \sin \beta$ and $\cos \alpha = \frac{9}{5 \sqrt{5}} \cos \beta$.
Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:
$(\frac{6}{5} \sin \beta)^2 + (\frac{9}{5 \sqrt{5}} \cos \beta)^2 = 1$
$\frac{36}{25} \sin^2 \beta + \frac{81}{125} \cos^2 \beta = 1$
Multiply by $125$:
$180 \sin^2 \beta + 81 \cos^2 \beta = 125$
Since $\cos^2 \beta = 1 - \sin^2 \beta$:
$180 \sin^2 \beta + 81(1 - \sin^2 \beta) = 125$
$180 \sin^2 \beta + 81 - 81 \sin^2 \beta = 125$
$99 \sin^2 \beta = 44$
$\sin^2 \beta = \frac{44}{99} = \frac{4}{9}$
$\sin \beta = \frac{2}{3}$ (since $\beta$ is acute).
Now,$\sin \alpha = \frac{6}{5} \sin \beta = \frac{6}{5} \times \frac{2}{3} = \frac{4}{5}$.

(C) Given $\sec \theta + \tan \theta = \frac{1}{3}$ . . . . . . $(i)$
Since $\sec^2 \theta - \tan^2 \theta = 1$,we have $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Thus,$\sec \theta - \tan \theta = 3$ . . . . . . $(ii)$
Adding $(i)$ and $(ii)$,$2 \sec \theta = \frac{1}{3} + 3 = \frac{10}{3} \Rightarrow \sec \theta = \frac{5}{3}$.
Subtracting $(ii)$ from $(i)$,$2 \tan \theta = \frac{1}{3} - 3 = \frac{-8}{3} \Rightarrow \tan \theta = \frac{-4}{3}$.
Since $\sec \theta > 0$ and $\tan \theta < 0$,$\theta$ lies in the $4^{th}$ quadrant.
Now,$\tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2(-4/3)}{1 - 16/9} = \frac{-8/3}{-7/9} = \frac{24}{7} > 0$.
Since $\sec \theta = 5/3$,$\cos \theta = 3/5$. Since $\theta$ is in the $4^{th}$ quadrant,$270^{\circ} < \theta < 360^{\circ}$.
Therefore,$540^{\circ} < 2 \theta < 720^{\circ}$.
Since $\tan 2 \theta > 0$ and $\sin 2 \theta = 2 \sin \theta \cos \theta = 2(-4/5)(3/5) = -24/25 < 0$,$2 \theta$ lies in the $3^{rd}$ quadrant (within the range $540^{\circ}$ to $720^{\circ}$).

(C) Given $\sinh x = \frac{\sqrt{21}}{2}$.
We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + \frac{21}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Using the identities $\sinh 2x = 2 \sinh x \cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$:
$\sinh 2x = 2 \times \frac{\sqrt{21}}{2} \times \frac{5}{2} = \frac{5\sqrt{21}}{2}$.
$\cosh 2x = (\frac{5}{2})^2 + (\frac{\sqrt{21}}{2})^2 = \frac{25}{4} + \frac{21}{4} = \frac{46}{4} = \frac{23}{2}$.
Therefore,$\cosh 2x + \sinh 2x = \frac{23}{2} + \frac{5\sqrt{21}}{2} = \frac{23 + 5\sqrt{21}}{2}$.

(B) Given expression: $\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}$
Substitute $\cot A = \frac{\cos A}{\sin A}$ and $\tan A = \frac{\sin A}{\cos A}$:
$= \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} + \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}$
$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} + \frac{\sin^2 A}{\cos A(\sin A-\cos A)}$
$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} - \frac{\sin^2 A}{\cos A(\cos A-\sin A)}$
$= \frac{1}{(\cos A-\sin A)} \left[ \frac{\cos^3 A - \sin^3 A}{\sin A \cos A} \right]$
Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(\cos A-\sin A)(\cos^2 A + \sin^2 A + \sin A \cos A)}{(\cos A-\sin A) \sin A \cos A}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A}$
$= \frac{1}{\sin A \cos A} + 1$
$= \operatorname{cosec} A \sec A + 1$

(A) Given expression: $\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A}$
Convert $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:
$= \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}$
$= \frac{1}{\sin A - \cos A} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right]$
Using $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$:
$= \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{(\sin A - \cos A)(\sin A \cos A)}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \operatorname{cosec} A + 1$

(C) Given,$P = \tan 15^{\circ} + \cot 15^{\circ}$,$Q = \tan 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{\circ}$.
For $P$: $P = \frac{\sin 15^{\circ}}{\cos 15^{\circ}} + \frac{\cos 15^{\circ}}{\sin 15^{\circ}} = \frac{\sin^2 15^{\circ} + \cos^2 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}} = \frac{1}{\frac{1}{2} \sin 30^{\circ}} = \frac{2}{1/2} = 4$.
For $Q$: $Q = \frac{\sin^2(22.5^{\circ}) + \cos^2(22.5^{\circ})}{\sin 22.5^{\circ} \cos 22.5^{\circ}} = \frac{1}{\frac{1}{2} \sin 45^{\circ}} = \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \approx 2.828$.
For $R$: $R = \sin 54^{\circ} + \sin 18^{\circ} = \cos 36^{\circ} + \sin 18^{\circ} = \frac{\sqrt{5}+1}{4} + \frac{\sqrt{5}-1}{4} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2} \approx 1.118$.
Comparing the values: $R \approx 1.118$,$Q \approx 2.828$,$P = 4$.
Thus,the ascending order is $R < Q < P$.

(A) $(I)$ The series is $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \dots + \sin^2 90^{\circ}$. There are $18$ terms from $5^{\circ}$ to $85^{\circ}$ plus $\sin^2 90^{\circ} = 1$. Pairing $\sin^2 \theta + \sin^2(90^{\circ}-\theta) = 1$,we have $8$ pairs plus $\sin^2 45^{\circ} = 0.5$ and $\sin^2 90^{\circ} = 1$. Total $= 8 + 0.5 + 1 = 9.5 = \frac{19}{2}$. Thus,$(I)$-$B$.
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 85^{\circ} = \tan^2 5^{\circ} \cdot \cot^2 5^{\circ} = 1$. There are $8$ such pairs and $\tan^2 45^{\circ} = 1$. Total $= 1^8 \cdot 1 = 1$. Thus,$(II)$-$D$.
$(III)$ $\cos^2 5^{\circ} + \dots + \cos^2 180^{\circ}$. Using $\cos^2 \theta + \cos^2(180^{\circ}-\theta) = 0$ is incorrect; rather $\cos(180^{\circ}-\theta) = -\cos \theta$,so $\cos^2(180^{\circ}-\theta) = \cos^2 \theta$. The sum is $\sum_{n=1}^{35} \cos^2(5n^{\circ}) + \cos^2 90^{\circ} + \cos^2 180^{\circ}$. This evaluates to $18$. Thus,$(III)$-$C$.
$(IV)$ $\cot \theta + \cot(180^{\circ}-\theta) = \cot \theta - \cot \theta = 0$. Pairing terms from $5^{\circ}$ to $175^{\circ}$ gives $0$. $\cot 90^{\circ} = 0$. Total $= 0$. Thus,$(IV)$-$A$.

(D) Given,$\sin \theta + \operatorname{cosec} \theta = 4$.
Squaring both sides,we get:
$(\sin \theta + \operatorname{cosec} \theta)^2 = 4^2$
$\sin^2 \theta + \operatorname{cosec}^2 \theta + 2 \sin \theta \operatorname{cosec} \theta = 16$
Since $\sin \theta \operatorname{cosec} \theta = 1$,we have:
$\sin^2 \theta + \operatorname{cosec}^2 \theta + 2(1) = 16$
$\sin^2 \theta + \operatorname{cosec}^2 \theta = 16 - 2 = 14$.

(D) Given expression: $\sqrt{\sin ^4 x+4 \cos ^2 x}-\sqrt{\cos ^4 x+4 \sin ^2 x}$
Using the identity $\cos ^2 x = 1-\sin ^2 x$ and $\sin ^2 x = 1-\cos ^2 x$:
$\sqrt{\sin ^4 x+4(1-\sin ^2 x)}-\sqrt{\cos ^4 x+4(1-\cos ^2 x)}$
$= \sqrt{\sin ^4 x-4 \sin ^2 x+4}-\sqrt{\cos ^4 x-4 \cos ^2 x+4}$
$= \sqrt{(\sin ^2 x-2)^2}-\sqrt{(\cos ^2 x-2)^2}$
Since $0 \le \sin ^2 x \le 1$ and $0 \le \cos ^2 x \le 1$,the terms inside the square roots are negative,so we take the absolute value:
$= |\sin ^2 x-2|-|\cos ^2 x-2|$
$= (2-\sin ^2 x)-(2-\cos ^2 x)$
$= \cos ^2 x-\sin ^2 x$
$= \cos 2 x$

(D) Given equation: $\cosh x - \frac{4}{5} \sinh x = 1$
Multiply by $5$: $5 \cosh x - 4 \sinh x = 5$
Substitute $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$:
$5 \left(\frac{e^x + e^{-x}}{2}\right) - 4 \left(\frac{e^x - e^{-x}}{2}\right) = 5$
Multiply by $2$: $5(e^x + e^{-x}) - 4(e^x - e^{-x}) = 10$
$5e^x + 5e^{-x} - 4e^x + 4e^{-x} = 10$
$e^x + 9e^{-x} = 10$
Multiply by $e^x$: $(e^x)^2 - 10e^x + 9 = 0$
$(e^x - 1)(e^x - 9) = 0$
Case $1$: $e^x = 1 \Rightarrow x = 0$
Case $2$: $e^x = 9 \Rightarrow x = \ln 9 = \ln(3^2) = 2 \ln 3$
Thus,the other solution is $x = 2 \log 3$.

(B) We know the fundamental identity for hyperbolic functions: $\cosh^2 u - \sinh^2 u = 1$.
Given that $\sinh u = \tan \theta$.
Substituting this into the identity,we get: $\cosh^2 u - (\tan \theta)^2 = 1$.
$\cosh^2 u = 1 + \tan^2 \theta$.
Using the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$,we have: $\cosh^2 u = \sec^2 \theta$.
Therefore,$\cosh u = \sec \theta$ (since $\cosh u$ is always positive).

(B) Given $\cos \theta = -\frac{\sqrt{3}}{2} < 0$. Since $\theta$ does not lie in the third quadrant,$\theta$ must lie in the second quadrant.
In the second quadrant,$\tan \theta = -\frac{1}{\sqrt{3}}$ and $\cot \theta = -\sqrt{3}$.
Thus,$\cot^2 \theta = (-\sqrt{3})^2 = 3$.
Given $\sin \alpha = -\frac{3}{5}$. Since $\sin \alpha < 0$,$\alpha$ lies in the third or fourth quadrant.
Case $1$: If $\alpha$ is in the third quadrant,$\tan \alpha = \frac{3}{4}$ and $\cos \alpha = -\frac{4}{5}$.
The expression becomes $\frac{2(\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}})}{3 - \frac{4}{5}} = \frac{\frac{3}{2} - 1}{\frac{15-4}{5}} = \frac{1/2}{11/5} = \frac{5}{22}$.
Case $2$: If $\alpha$ is in the fourth quadrant,$\tan \alpha = -\frac{3}{4}$ and $\cos \alpha = \frac{4}{5}$.
The expression becomes $\frac{2(-\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}})}{3 + \frac{4}{5}} = \frac{-\frac{3}{2} - 1}{\frac{15+4}{5}} = \frac{-5/2}{19/5} = -\frac{25}{38}$.
Since $\frac{5}{22}$ is the only option provided,the correct answer is $\frac{5}{22}$.

(A) Given,$\sin \theta + \operatorname{cosec} \theta = 2$.
We know that $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$.
So,$\sin \theta + \frac{1}{\sin \theta} = 2$.
Let $\sin \theta = x$,then $x + \frac{1}{x} = 2$,which implies $x^2 - 2x + 1 = 0$,or $(x - 1)^2 = 0$.
Thus,$x = 1$,which means $\sin \theta = 1$.
Consequently,$\operatorname{cosec} \theta = 1$.
Therefore,$\sin^{10} \theta + \operatorname{cosec}^{10} \theta = (1)^{10} + (1)^{10} = 1 + 1 = 2$.

(B) Given $0 < \theta < \frac{\pi}{2}$ and $\sin \theta \cos \theta = \frac{12}{25}$.
We need to find $\sin^4 \theta + \cos^4 \theta$.
Using the identity $a^2 + b^2 = (a + b)^2 - 2ab$,we have:
$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta)^2 + (\cos^2 \theta)^2$
$= (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$= (1)^2 - 2(\sin \theta \cos \theta)^2$
$= 1 - 2 \left(\frac{12}{25}\right)^2$
$= 1 - 2 \left(\frac{144}{625}\right)$
$= 1 - \frac{288}{625}$
$= \frac{625 - 288}{625} = \frac{337}{625}$.

(D) Given $\sin(\theta) + \operatorname{cosec}(\theta) = 2$.
Since $\operatorname{cosec}(\theta) = \frac{1}{\sin(\theta)}$,we have $\sin(\theta) + \frac{1}{\sin(\theta)} = 2$.
Let $x = \sin(\theta)$,then $x + \frac{1}{x} = 2$,which implies $x^2 - 2x + 1 = 0$,so $(x - 1)^2 = 0$.
Thus,$\sin(\theta) = 1$.
Therefore,$\sin^{2020}(\theta) + \operatorname{cosec}^{2020}(\theta) = (1)^{2020} + (1)^{2020} = 1 + 1 = 2$.

(B) Given,$\sec \theta = m$ and $\tan \theta = n$.
We know that $\sec^2 \theta - \tan^2 \theta = 1$,so $m^2 - n^2 = 1$,which implies $(m - n)(m + n) = 1$.
Therefore,$\frac{1}{m + n} = m - n$.
Now,substitute this into the expression:
$\frac{1}{m} [m + n + (m - n)] = \frac{1}{m} [2m] = 2$.

(C) Given expression: $\frac{1-\cos(2x)+\sin(x)}{\sin(2x)+\cos(x)}$
Using the identity $\cos(2x) = 1 - 2\sin^2(x)$,the numerator becomes: $1 - (1 - 2\sin^2(x)) + \sin(x) = 2\sin^2(x) + \sin(x) = \sin(x)(2\sin(x) + 1)$
Using the identity $\sin(2x) = 2\sin(x)\cos(x)$,the denominator becomes: $2\sin(x)\cos(x) + \cos(x) = \cos(x)(2\sin(x) + 1)$
Substituting these back into the expression: $\frac{\sin(x)(2\sin(x) + 1)}{\cos(x)(2\sin(x) + 1)} = \frac{\sin(x)}{\cos(x)} = \tan(x)$

(A) Given,$4 \cos x + 3 \sin x = 5$.
Divide both sides by $\cos x$ (assuming $\cos x \neq 0$):
$4 + 3 \tan x = 5 \sec x$.
Squaring both sides:
$(4 + 3 \tan x)^2 = (5 \sec x)^2$.
$16 + 9 \tan^2 x + 24 \tan x = 25 \sec^2 x$.
Using the identity $\sec^2 x = 1 + \tan^2 x$:
$16 + 9 \tan^2 x + 24 \tan x = 25(1 + \tan^2 x)$.
$16 + 9 \tan^2 x + 24 \tan x = 25 + 25 \tan^2 x$.
Rearranging the terms:
$16 \tan^2 x - 24 \tan x + 9 = 0$.
$(4 \tan x - 3)^2 = 0$.
Therefore,$4 \tan x = 3$,which gives $\tan x = \frac{3}{4}$.

(D) Given that $\sec \theta + \tan \theta = 2/3$ $(i)$
We know that $\sec^2 \theta - \tan^2 \theta = 1$,which implies $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value from $(i)$,we get $\sec \theta - \tan \theta = 3/2$ (ii)
Adding $(i)$ and (ii): $2 \sec \theta = 2/3 + 3/2 = (4 + 9)/6 = 13/6$,so $\sec \theta = 13/12$.
Subtracting (ii) from $(i)$: $2 \tan \theta = 2/3 - 3/2 = (4 - 9)/6 = -5/6$,so $\tan \theta = -5/12$.
Since $\sec \theta > 0$ and $\tan \theta < 0$,$\theta$ must lie in the $IV$ quadrant.