Trigonometrical ratios of multiple and sub-multiple angles Questions in English

(B) Given $\sin \theta = -\frac{4}{5}$ and $\theta$ lies in the $III$ quadrant.
Since $\theta$ is in the $III$ quadrant,$\cos \theta$ must be negative.
$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - (-\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.
We know that $\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$.
Since $\pi < \theta < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$.
This means $\frac{\theta}{2}$ lies in the $II$ quadrant,where cosine is negative.
Therefore,$\cos \frac{\theta}{2} = -\sqrt{\frac{1 + (-\frac{3}{5})}{2}} = -\sqrt{\frac{2/5}{2}} = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}}$.

(B) Given that $\cos \theta = \frac{1}{2}\left( x + \frac{1}{x} \right)$.
$\Rightarrow x + \frac{1}{x} = 2 \cos \theta$.
We know that $x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2$.
Substituting the value,we get $x^2 + \frac{1}{x^2} = (2 \cos \theta)^2 - 2 = 4 \cos^2 \theta - 2$.
Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we have $4 \cos^2 \theta - 2 = 2(2 \cos^2 \theta - 1) = 2 \cos 2\theta$.
Therefore,$\frac{1}{2}\left( x^2 + \frac{1}{x^2} \right) = \frac{1}{2} \times 2 \cos 2\theta = \cos 2\theta$.

(C) We know that $\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$.
$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}$
$= \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}$
Using the identities $\cos^2 x - \sin^2 x = \cos 2x$ and $2\sin x \cos x = \sin 2x$,we get:
$= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2\cos 2x}{\sin 2x} = 2\cot 2x$.

(C) Given expression: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}$
Using the identities $1 - \cos A = 2 \sin^2 \frac{A}{2}$,$1 + \cos A = 2 \cos^2 \frac{A}{2}$,and $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$:
$= \frac{2 \sin^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2}}$
$= \frac{2 \sin \frac{A}{2} (\sin \frac{A}{2} + \cos \frac{A}{2})}{2 \cos \frac{A}{2} (\cos \frac{A}{2} + \sin \frac{A}{2})}$
$= \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} = \tan \frac{A}{2}$
Trick: Put $A = 60^\circ$.
$\frac{1 + \sin 60^\circ - \cos 60^\circ}{1 + \sin 60^\circ + \cos 60^\circ} = \frac{1 + \frac{\sqrt{3}}{2} - \frac{1}{2}}{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}} = \frac{\frac{1}{2} + \frac{\sqrt{3}}{2}}{\frac{3}{2} + \frac{\sqrt{3}}{2}} = \frac{1 + \sqrt{3}}{3 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{3}(\sqrt{3} + 1)} = \frac{1}{\sqrt{3}}$
Since $\tan \frac{60^\circ}{2} = \tan 30^\circ = \frac{1}{\sqrt{3}}$,the correct option is $(C)$.

(A) We use the expansion formulas $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Given expression: $\tan \left( \frac{\pi }{4} + \theta \right) - \tan \left( \frac{\pi }{4} - \theta \right)$.
Substituting $\tan \frac{\pi }{4} = 1$,we get:
$= \frac{1 + \tan \theta}{1 - \tan \theta} - \frac{1 - \tan \theta}{1 + \tan \theta}$.
Taking the common denominator:
$= \frac{(1 + \tan \theta)^2 - (1 - \tan \theta)^2}{(1 - \tan \theta)(1 + \tan \theta)}$.
Using $(a+b)^2 - (a-b)^2 = 4ab$ and $(a-b)(a+b) = a^2 - b^2$:
$= \frac{4 \tan \theta}{1 - \tan^2 \theta}$.
$= 2 \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)$.
Since $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,the expression simplifies to:
$= 2 \tan 2\theta$.

(C) Given the expression: $\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right) = 3$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan \left( \frac{\pi}{3} + x \right) = \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x}$ and $\tan \left( \frac{2\pi}{3} + x \right) = \frac{-\sqrt{3} + \tan x}{1 + \sqrt{3} \tan x}$.
Summing these: $\tan x + \frac{\sqrt{3} + \tan x}{1 - \sqrt{3} \tan x} + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} = \tan x + \frac{(\sqrt{3} + \tan x)(1 + \sqrt{3} \tan x) + (\tan x - \sqrt{3})(1 - \sqrt{3} \tan x)}{1 - 3 \tan^2 x}$.
$= \tan x + \frac{\sqrt{3} + 3 \tan x + \tan x + \sqrt{3} \tan^2 x + \tan x - \sqrt{3} \tan^2 x - \sqrt{3} + 3 \tan x}{1 - 3 \tan^2 x} = \tan x + \frac{8 \tan x}{1 - 3 \tan^2 x}$.
$= \frac{\tan x - 3 \tan^3 x + 8 \tan x}{1 - 3 \tan^2 x} = \frac{9 \tan x - 3 \tan^3 x}{1 - 3 \tan^2 x} = 3 \left( \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \right) = 3 \tan 3x$.
Given $3 \tan 3x = 3$,we get $\tan 3x = 1$.

(D) We use the formula $\prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Here,$\theta = \frac{\pi}{5}$ and $n = 4$.
$\cos \frac{\pi}{5} \cos \frac{2\pi}{5} \cos \frac{4\pi}{5} \cos \frac{8\pi}{5} = \frac{\sin(2^4 \cdot \frac{\pi}{5})}{2^4 \sin \frac{\pi}{5}} = \frac{\sin \frac{16\pi}{5}}{16 \sin \frac{\pi}{5}}$.
Since $\sin \frac{16\pi}{5} = \sin(3\pi + \frac{\pi}{5}) = -\sin \frac{\pi}{5}$,
The expression becomes $\frac{-\sin \frac{\pi}{5}}{16 \sin \frac{\pi}{5}} = -\frac{1}{16}$.

(C) We use the identity $\tan \theta \tan(60^\circ - \theta) \tan(60^\circ + \theta) = \tan 3\theta$.
Given expression is $E = \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ$.
Rearranging the terms,we have $E = \tan 60^\circ (\tan 20^\circ \tan 40^\circ \tan 80^\circ)$.
Note that $\tan 80^\circ = \tan(60^\circ + 20^\circ)$ and $\tan 40^\circ = \tan(60^\circ - 20^\circ)$.
Using the identity for $\theta = 20^\circ$:
$\tan 20^\circ \tan(60^\circ - 20^\circ) \tan(60^\circ + 20^\circ) = \tan(3 \times 20^\circ) = \tan 60^\circ$.
Substituting this back into the expression for $E$:
$E = \tan 60^\circ \times \tan 60^\circ = \sqrt{3} \times \sqrt{3} = 3$.

(D) We use the formula $\cos \theta \cos(2\theta) \cos(4\theta) = \frac{\sin(8\theta)}{8\sin \theta}$.
Here,$\theta = 20^\circ$.
So,$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin(8 \times 20^\circ)}{8 \sin 20^\circ}$.
$= \frac{\sin 160^\circ}{8 \sin 20^\circ}$.
Since $\sin 160^\circ = \sin(180^\circ - 20^\circ) = \sin 20^\circ$,
$= \frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$.

(B) Given $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ$.
Multiply and divide by $2 \sin 10^\circ$:
$x = \frac{2 \sin 10^\circ \cos 10^\circ \cos 20^\circ \cos 40^\circ}{2 \sin 10^\circ}$
Using $2 \sin \theta \cos \theta = \sin 2\theta$:
$x = \frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ}{2 \sin 10^\circ} = \frac{2 \sin 20^\circ \cos 20^\circ \cos 40^\circ}{4 \sin 10^\circ}$
$x = \frac{\sin 40^\circ \cos 40^\circ}{4 \sin 10^\circ} = \frac{2 \sin 40^\circ \cos 40^\circ}{8 \sin 10^\circ}$
$x = \frac{\sin 80^\circ}{8 \sin 10^\circ}$
Since $\sin 80^\circ = \cos 10^\circ$:
$x = \frac{\cos 10^\circ}{8 \sin 10^\circ} = \frac{1}{8} \cot 10^\circ$.

(A) We know that $3A = 2A + A$.
Taking tangent on both sides: $\tan 3A = \tan(2A + A)$.
Using the formula $\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\tan 3A = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}$.
Cross-multiplying:
$\tan 3A(1 - \tan 2A \tan A) = \tan 2A + \tan A$.
$\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A$.
Rearranging the terms:
$\tan 3A - \tan 2A - \tan A = \tan 3A \tan 2A \tan A$.

(A) Given expression: $2\cos x - (\cos 5x + \cos 3x)$
Using the formula $\cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$:
$\cos 5x + \cos 3x = 2\cos(4x)\cos(x)$
Substituting this into the expression:
$2\cos x - 2\cos 4x \cos x = 2\cos x(1 - \cos 4x)$
Using the identity $1 - \cos 2\theta = 2\sin^2 \theta$,where $2\theta = 4x$ (so $\theta = 2x$):
$2\cos x(2\sin^2 2x) = 4\cos x \sin^2 2x$
Using $\sin 2x = 2\sin x \cos x$,so $\sin^2 2x = 4\sin^2 x \cos^2 x$:
$4\cos x(4\sin^2 x \cos^2 x) = 16\cos^3 x \sin^2 x$
Thus,the correct option is $A$.

(D) We use the formula $\cos \theta \cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Here,$\theta = \frac{\pi}{7}$ and $n = 3$.
Thus,$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{\sin(2^3 \cdot \frac{\pi}{7})}{2^3 \sin(\frac{\pi}{7})}$.
$= \frac{\sin(\frac{8\pi}{7})}{8 \sin(\frac{\pi}{7})}$.
Since $\sin(\frac{8\pi}{7}) = \sin(\pi + \frac{\pi}{7}) = -\sin(\frac{\pi}{7})$.
$= \frac{-\sin(\frac{\pi}{7})}{8 \sin(\frac{\pi}{7})} = -\frac{1}{8}$.

(A) Given that $\sec \theta = 1\frac{1}{4} = \frac{5}{4}$.
We use the identity $\sec \theta = \frac{1 + \tan^2(\theta/2)}{1 - \tan^2(\theta/2)}$.
Substituting the value of $\sec \theta$:
$\frac{5}{4} = \frac{1 + \tan^2(\theta/2)}{1 - \tan^2(\theta/2)}$.
Cross-multiplying gives:
$5(1 - \tan^2(\theta/2)) = 4(1 + \tan^2(\theta/2))$.
$5 - 5\tan^2(\theta/2) = 4 + 4\tan^2(\theta/2)$.
Rearranging the terms:
$5 - 4 = 4\tan^2(\theta/2) + 5\tan^2(\theta/2)$.
$1 = 9\tan^2(\theta/2)$.
$\tan^2(\theta/2) = \frac{1}{9}$.
Taking the square root,we get $\tan(\theta/2) = \frac{1}{3}$ (assuming $\theta$ is in the first quadrant).

(D) Given that $\tan \frac{A}{2} = \frac{3}{2}$.
Using the trigonometric identities $1 + \cos A = 2 \cos^2 \frac{A}{2}$ and $1 - \cos A = 2 \sin^2 \frac{A}{2}$,we have:
$\frac{1 + \cos A}{1 - \cos A} = \frac{2 \cos^2 \frac{A}{2}}{2 \sin^2 \frac{A}{2}}$
$= \cot^2 \frac{A}{2}$
$= \left( \frac{1}{\tan \frac{A}{2}} \right)^2 = \left( \frac{1}{3/2} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

(D) Given that $\cos A = \frac{\sqrt{3}}{2}$.
Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,we have $A = 30^\circ$.
Now,calculate $\tan 3A$:
$\tan 3A = \tan(3 \times 30^\circ) = \tan 90^\circ$.
Since $\tan 90^\circ$ is undefined,the value is $\infty$.

(A) We know that $\sin 4\theta = 2\sin 2\theta \cos 2\theta$.
Using the double angle formulas $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\theta = 1 - 2\sin^2 \theta$,we get:
$\sin 4\theta = 2(2\sin \theta \cos \theta)(1 - 2\sin^2 \theta)$.
Since $\cos \theta = \sqrt{1 - \sin^2 \theta}$ (assuming $\cos \theta > 0$),we can write:
$\sin 4\theta = 4\sin \theta (1 - 2\sin^2 \theta)\sqrt{1 - \sin^2 \theta}$.

(A) Given that $\tan \theta = \frac{b}{a}$.
Now,$a \cos 2\theta + b \sin 2\theta = a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$.
Substituting $\tan \theta = \frac{b}{a}$,we get:
$= a \left( \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} \right) + b \left( \frac{2 \frac{b}{a}}{1 + \frac{b^2}{a^2}} \right)$
$= a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2ab}{a^2 + b^2} \right)$
$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2}$
$= \frac{a^3 + ab^2}{a^2 + b^2} = \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

(A) We are given the expression: $\left( \frac{\sin 2A}{1 + \cos 2A} \right) \left( \frac{\cos A}{1 + \cos A} \right)$
Using the double angle identities $\sin 2A = 2 \sin A \cos A$ and $1 + \cos 2A = 2 \cos^2 A$,we get:
$= \left( \frac{2 \sin A \cos A}{2 \cos^2 A} \right) \left( \frac{\cos A}{1 + \cos A} \right)$
$= \left( \frac{\sin A}{\cos A} \right) \left( \frac{\cos A}{1 + \cos A} \right)$
$= \frac{\sin A}{1 + \cos A}$
Using the half-angle identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$ and $1 + \cos A = 2 \cos^2 \frac{A}{2}$:
$= \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos^2 \frac{A}{2}}$
$= \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} = \tan \frac{A}{2}$.

(D) Given expression: $\frac{1}{\tan 3A - \tan A} - \frac{1}{\cot 3A - \cot A}$
Since $\cot \theta = \frac{1}{\tan \theta}$,we have $\frac{1}{\cot 3A - \cot A} = \frac{1}{\frac{1}{\tan 3A} - \frac{1}{\tan A}} = \frac{\tan A \tan 3A}{\tan A - \tan 3A} = -\frac{\tan A \tan 3A}{\tan 3A - \tan A}$.
Substituting this back into the expression:
$= \frac{1}{\tan 3A - \tan A} - (-\frac{\tan A \tan 3A}{\tan 3A - \tan A})$
$= \frac{1 + \tan A \tan 3A}{\tan 3A - \tan A}$
Using the identity $\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$,we know that $\tan(3A - A) = \frac{\tan 3A - \tan A}{1 + \tan 3A \tan A}$.
Therefore,$\frac{1 + \tan 3A \tan A}{\tan 3A - \tan A} = \frac{1}{\tan(3A - A)} = \frac{1}{\tan 2A} = \cot 2A$.

(A) Given expression: $\csc A - 2 \cot 2A \cos A$
$= \frac{1}{\sin A} - \frac{2 \cos A \cos 2A}{\sin 2A}$
$= \frac{1}{\sin A} - \frac{2 \cos A \cos 2A}{2 \sin A \cos A}$
$= \frac{1}{\sin A} - \frac{\cos 2A}{\sin A}$
$= \frac{1 - \cos 2A}{\sin A}$
Using the identity $1 - \cos 2A = 2 \sin^2 A$:
$= \frac{2 \sin^2 A}{\sin A}$
$= 2 \sin A$

(A) Given expression: $\sqrt{2 + \sqrt{2 + 2\cos 4\theta}}$
Using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta$.
Substituting this into the expression:
$= \sqrt{2 + \sqrt{4\cos^2 2\theta}}$
$= \sqrt{2 + 2\cos 2\theta}$
Again,using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 2\theta = 2(1 + \cos 2\theta) = 2(2\cos^2 \theta) = 4\cos^2 \theta$.
$= \sqrt{4\cos^2 \theta}$
$= 2\cos \theta$.

(C) We know the trigonometric identity for $\cos 3\theta$ is:
$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$
Comparing this with the given equation $\cos 3\theta = \alpha \cos \theta + \beta \cos^3 \theta$,we get:
$\alpha = -3$ and $\beta = 4$
Therefore,$(\alpha, \beta) = (-3, 4)$.

(B) We know the formula for $\tan 3A$ is $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$.
Given $\tan A = \frac{1}{2}$.
Substituting the value of $\tan A$ in the formula:
$\tan 3A = \frac{3(\frac{1}{2}) - (\frac{1}{2})^3}{1 - 3(\frac{1}{2})^2}$
$= \frac{\frac{3}{2} - \frac{1}{8}}{1 - 3(\frac{1}{4})}$
$= \frac{\frac{12-1}{8}}{1 - \frac{3}{4}}$
$= \frac{\frac{11}{8}}{\frac{1}{4}}$
$= \frac{11}{8} \times 4 = \frac{11}{2}$.

(B) Given expression: $E = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}$
Since $1 \pm \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \pm 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} \pm \sin \frac{x}{2})^2$,we have $\sqrt{1 \pm \sin x} = |\cos \frac{x}{2} \pm \sin \frac{x}{2}|$.
For $x$ in the $II^{nd}$ quadrant,$\frac{\pi}{2} < x < \pi$,so $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
In this interval,$\cos \frac{x}{2} > 0$ and $\sin \frac{x}{2} > 0$,with $\cos \frac{x}{2} > \sin \frac{x}{2}$.
Thus,$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.
Substituting these into the expression: $E = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}$
$E = \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}$.
Note: The provided option $(b)$ $\tan \frac{x}{2}$ is incorrect based on the quadrant analysis. The correct result is $\cot \frac{x}{2}$.

(D) Given expression: $(\sec 2A + 1)\sec^2 A$
Substitute $\sec 2A = \frac{1}{\cos 2A} = \frac{1+\tan^2 A}{1-\tan^2 A}$ and $\sec^2 A = 1+\tan^2 A$:
$= \left( \frac{1+\tan^2 A}{1-\tan^2 A} + 1 \right)(1+\tan^2 A)$
$= \left( \frac{1+\tan^2 A + 1 - \tan^2 A}{1-\tan^2 A} \right)(1+\tan^2 A)$
$= \left( \frac{2}{1-\tan^2 A} \right)(1+\tan^2 A)$
$= \frac{2(1+\tan^2 A)}{1-\tan^2 A} = 2 \sec 2A$

(B) Given expression: $2\sin A\cos^3 A - 2\sin^3 A\cos A$
Factor out $2\sin A\cos A$:
$= 2\sin A\cos A(\cos^2 A - \sin^2 A)$
Using the identities $\sin 2A = 2\sin A\cos A$ and $\cos 2A = \cos^2 A - \sin^2 A$:
$= \sin 2A \cos 2A$
Multiply and divide by $2$:
$= \frac{1}{2}(2\sin 2A \cos 2A)$
Using the identity $\sin 2\theta = 2\sin \theta \cos \theta$ where $\theta = 2A$:
$= \frac{1}{2}\sin 4A$

(C) Given expression: $\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta}$
Using the identities $\sin 2\theta = 2 \sin \theta \cos \theta$ and $\cos 2\theta = 2 \cos^2 \theta - 1$:
Numerator: $\sin \theta + 2 \sin \theta \cos \theta = \sin \theta (1 + 2 \cos \theta)$
Denominator: $1 + \cos \theta + (2 \cos^2 \theta - 1) = \cos \theta + 2 \cos^2 \theta = \cos \theta (1 + 2 \cos \theta)$
Substituting these back into the expression:
$\frac{\sin \theta (1 + 2 \cos \theta)}{\cos \theta (1 + 2 \cos \theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta$
Thus,the correct option is $C$.

(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$.
Using half-angle identities $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ and $1 + \cos \alpha = 2\cos^2 \frac{\alpha}{2}$:
$\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2})}{2\cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = y$
$\frac{4\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2\cos \frac{\alpha}{2}(\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2})} = y$
$\frac{2\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}} = y$.
Now,consider the expression $E = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha}$.
Using $1 - \cos \alpha = 2\sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$ and $1 + \sin \alpha = (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2$:
$E = \frac{2\sin^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}} = y$.

(A) Given $\sin \beta = \frac{1}{\sqrt{10}}$.
Since $0 < \beta < \frac{\pi}{2}$,we have $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Thus,$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$.
Now,$\tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Consider $\tan(\alpha + 2\beta) = \frac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$.
Since $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$,we have $0 < 2\beta < \frac{\pi}{2}$ (as $\tan 2\beta = 3/4 < 1$).
Therefore,$0 < \alpha + 2\beta < \pi$. Since $\tan(\alpha + 2\beta) = 1$,we must have $\alpha + 2\beta = \frac{\pi}{4}$.
Thus,$2\beta = \frac{\pi}{4} - \alpha$.

(A) Given $\tan A = \frac{1 - \cos B}{\sin B}$.
Using the half-angle formulas $1 - \cos B = 2\sin^2(B/2)$ and $\sin B = 2\sin(B/2)\cos(B/2)$:
$\tan A = \frac{2\sin^2(B/2)}{2\sin(B/2)\cos(B/2)} = \frac{\sin(B/2)}{\cos(B/2)} = \tan(B/2)$.
Thus,$A = B/2$,which implies $2A = B$.
Therefore,$\tan 2A = \tan B$.

(B) Given expression: $\frac{\sec 8A - 1}{\sec 4A - 1}$
$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} \times \frac{\cos 4A}{1 - \cos 4A}$
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$:
$= \frac{2 \sin^2 4A}{\cos 8A} \times \frac{\cos 4A}{2 \sin^2 2A}$
$= \frac{(2 \sin 4A \cos 4A) \sin 4A}{\cos 8A \times 2 \sin^2 2A}$
Since $2 \sin 4A \cos 4A = \sin 8A$:
$= \frac{\sin 8A}{\cos 8A} \times \frac{\sin 4A}{2 \sin^2 2A} = \tan 8A \times \frac{2 \sin 2A \cos 2A}{2 \sin^2 2A}$
$= \tan 8A \times \frac{\cos 2A}{\sin 2A} = \tan 8A \times \cot 2A = \frac{\tan 8A}{\tan 2A}$

(C) We use the formula $2\sin X \sin Y = \cos(X - Y) - \cos(X + Y)$.
Given expression: $32\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2} \right) = 16 \times 2\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2} \right)$.
$= 16 \left[ \cos\left( \frac{5A}{2} - \frac{A}{2} \right) - \cos\left( \frac{5A}{2} + \frac{A}{2} \right) \right]$.
$= 16 (\cos 2A - \cos 3A)$.
Using $\cos 2A = 2\cos^2 A - 1$ and $\cos 3A = 4\cos^3 A - 3\cos A$:
$= 16 [ (2\cos^2 A - 1) - (4\cos^3 A - 3\cos A) ]$.
Given $\cos A = \frac{3}{4}$,so $\cos^2 A = \frac{9}{16}$ and $\cos^3 A = \frac{27}{64}$.
$= 16 \left[ \left( 2 \times \frac{9}{16} - 1 \right) - \left( 4 \times \frac{27}{64} - 3 \times \frac{3}{4} \right) \right]$.
$= 16 \left[ \left( \frac{9}{8} - 1 \right) - \left( \frac{27}{16} - \frac{9}{4} \right) \right]$.
$= 16 \left[ \frac{1}{8} - \left( \frac{27 - 36}{16} \right) \right] = 16 \left[ \frac{1}{8} - \left( -\frac{9}{16} \right) \right]$.
$= 16 \left( \frac{2 + 9}{16} \right) = 11$.

(B) Given $\tan \alpha = \frac{1}{7}$.
Using the formula $\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}$,we get:
$\cos 2\alpha = \frac{1 - (1/7)^2}{1 + (1/7)^2} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.
Now,given $\tan \beta = \frac{1}{3}$.
Using the formula $\sin 2\beta = \frac{2 \tan \beta}{1 + \tan^2 \beta}$,we get:
$\sin 2\beta = \frac{2(1/3)}{1 + (1/3)^2} = \frac{2/3}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2}{3} \times \frac{9}{10} = \frac{3}{5}$.
Since $\sin 2\beta = \frac{3}{5}$,then $\cos 2\beta = \sqrt{1 - (3/5)^2} = \frac{4}{5}$.
Now,calculate $\sin 4\beta = 2 \sin 2\beta \cos 2\beta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$.
Thus,$\cos 2\alpha = \sin 4\beta$.

(B) Given $\cos A = \frac{3}{4}$.
We use the product-to-sum formula: $2\sin X \cos Y = \sin(X+Y) + \sin(X-Y)$.
$32\sin \frac{A}{2}\cos \frac{5A}{2} = 16(2\sin \frac{A}{2}\cos \frac{5A}{2})$
$= 16(\sin(\frac{A}{2} + \frac{5A}{2}) + \sin(\frac{A}{2} - \frac{5A}{2}))$
$= 16(\sin(3A) + \sin(-2A))$
$= 16(\sin 3A - \sin 2A)$
$= 16((3\sin A - 4\sin^3 A) - 2\sin A \cos A)$
$= 16\sin A(3 - 4\sin^2 A - 2\cos A)$
Since $\cos A = \frac{3}{4}$,$\sin^2 A = 1 - \cos^2 A = 1 - \frac{9}{16} = \frac{7}{16}$,so $\sin A = \frac{\sqrt{7}}{4}$.
$= 16 \times \frac{\sqrt{7}}{4} \times (3 - 4(\frac{7}{16}) - 2(\frac{3}{4}))$
$= 4\sqrt{7} \times (3 - \frac{7}{4} - \frac{3}{2})$
$= 4\sqrt{7} \times (\frac{12 - 7 - 6}{4})$
$= 4\sqrt{7} \times (-\frac{1}{4}) = -\sqrt{7}$.

(D) We have $\frac{\cos A}{1 - \sin A}$.
Multiply the numerator and denominator by $(1 + \sin A)$:
$\frac{\cos A(1 + \sin A)}{(1 - \sin A)(1 + \sin A)} = \frac{\cos A(1 + \sin A)}{1 - \sin^2 A} = \frac{\cos A(1 + \sin A)}{\cos^2 A} = \frac{1 + \sin A}{\cos A}$.
Using half-angle identities $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$ and $\cos A = \cos^2 \frac{A}{2} - \sin^2 \frac{A}{2}$:
$= \frac{(\cos \frac{A}{2} + \sin \frac{A}{2})^2}{(\cos \frac{A}{2} - \sin \frac{A}{2})(\cos \frac{A}{2} + \sin \frac{A}{2})} = \frac{\cos \frac{A}{2} + \sin \frac{A}{2}}{\cos \frac{A}{2} - \sin \frac{A}{2}}$.
Dividing numerator and denominator by $\cos \frac{A}{2}$:
$= \frac{1 + \tan \frac{A}{2}}{1 - \tan \frac{A}{2}} = \tan \left( \frac{\pi}{4} + \frac{A}{2} \right)$.

(C) We know that $\tan \frac{A}{2} = \frac{\sin(A/2)}{\cos(A/2)}$.
Using the half-angle formulas,$\sin(A/2) = \pm \sqrt{\frac{1 - \cos A}{2}}$ and $\cos(A/2) = \pm \sqrt{\frac{1 + \cos A}{2}}$.
Therefore,$\tan \frac{A}{2} = \frac{\pm \sqrt{(1 - \cos A)/2}}{\pm \sqrt{(1 + \cos A)/2}} = \pm \sqrt{\frac{1 - \cos A}{1 + \cos A}}$.

(A) Given $\sin \alpha = \frac{-3}{5}$ and $\pi < \alpha < \frac{3\pi}{2}$ (i.e.,$\alpha$ is in the $III^{rd}$ quadrant).
Since $\alpha$ is in the $III^{rd}$ quadrant,$\cos \alpha$ must be negative.
$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - (\frac{-3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Since $\pi < \alpha < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4}$.
This means $\frac{\alpha}{2}$ lies in the $II^{nd}$ quadrant,where $\cos$ is negative.
Using the formula $\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + \cos \alpha}{2}}$:
$\cos \frac{\alpha}{2} = -\sqrt{\frac{1 + (-4/5)}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(B) Given expression: $\sec 2x - \tan 2x = \frac{1 - \sin 2x}{\cos 2x}$
Using identities $\sin 2x = 2 \sin x \cos x$,$\cos 2x = \cos^2 x - \sin^2 x$,and $1 = \cos^2 x + \sin^2 x$:
$= \frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{(\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$
$= \frac{\cos x - \sin x}{\cos x + \sin x}$
Dividing numerator and denominator by $\cos x$:
$= \frac{1 - \tan x}{1 + \tan x} = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x}$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$= \tan \left( \frac{\pi}{4} - x \right)$.

(A) Given $\tan \theta = t.$
We know that $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2t}{1 - t^2}$ and $\sec 2\theta = \frac{1}{\cos 2\theta} = \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} = \frac{1 + t^2}{1 - t^2}.$
Therefore,$\tan 2\theta + \sec 2\theta = \frac{2t}{1 - t^2} + \frac{1 + t^2}{1 - t^2}.$
$= \frac{2t + 1 + t^2}{1 - t^2} = \frac{(1 + t)^2}{(1 - t)(1 + t)}.$
$= \frac{1 + t}{1 - t}.$

(C) Given $\cos \theta = \frac{1}{2}\left( a + \frac{1}{a} \right).$
Using the identity $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta,$
$\cos 3\theta = 4 \left[ \frac{1}{2} \left( a + \frac{1}{a} \right) \right]^3 - 3 \left[ \frac{1}{2} \left( a + \frac{1}{a} \right) \right]$
$\cos 3\theta = 4 \cdot \frac{1}{8} \left( a + \frac{1}{a} \right)^3 - \frac{3}{2} \left( a + \frac{1}{a} \right)$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left[ \left( a + \frac{1}{a} \right)^2 - 3 \right]$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left[ a^2 + 2 + \frac{1}{a^2} - 3 \right]$
$\cos 3\theta = \frac{1}{2} \left( a + \frac{1}{a} \right) \left( a^2 - 1 + \frac{1}{a^2} \right)$
$\cos 3\theta = \frac{1}{2} \left( a^3 - a + \frac{1}{a} + a - \frac{1}{a} + \frac{1}{a^3} \right)$
$\cos 3\theta = \frac{1}{2} \left( a^3 + \frac{1}{a^3} \right).$

(C) Given $A = 133^\circ$,so $\frac{A}{2} = 66.5^\circ$.
Since $45^\circ < \frac{A}{2} < 90^\circ$,we have $\sin \frac{A}{2} > \cos \frac{A}{2} > 0$.
We know that $1 + \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} + 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} + \cos \frac{A}{2})^2$.
Thus,$\sqrt{1 + \sin A} = |\sin \frac{A}{2} + \cos \frac{A}{2}| = \sin \frac{A}{2} + \cos \frac{A}{2} \dots (i)$.
Similarly,$1 - \sin A = \sin^2 \frac{A}{2} + \cos^2 \frac{A}{2} - 2\sin \frac{A}{2}\cos \frac{A}{2} = (\sin \frac{A}{2} - \cos \frac{A}{2})^2$.
Thus,$\sqrt{1 - \sin A} = |\sin \frac{A}{2} - \cos \frac{A}{2}|$.
Since $\sin \frac{A}{2} > \cos \frac{A}{2}$,we have $\sqrt{1 - \sin A} = \sin \frac{A}{2} - \cos \frac{A}{2} \dots (ii)$.
Subtracting $(ii)$ from $(i)$:
$\sqrt{1 + \sin A} - \sqrt{1 - \sin A} = (\sin \frac{A}{2} + \cos \frac{A}{2}) - (\sin \frac{A}{2} - \cos \frac{A}{2}) = 2\cos \frac{A}{2}$.

(D) Given $\sin A = \frac{4}{5}$ and $90^\circ < A < 180^\circ$ (second quadrant).
Since $\sin A$ is positive and $\cos A$ is negative in the second quadrant,$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (\frac{4}{5})^2} = -\frac{3}{5}$.
We know that $\tan \frac{A}{2} = \sqrt{\frac{1 - \cos A}{1 + \cos A}}$.
Since $90^\circ < A < 180^\circ$,we have $45^\circ < \frac{A}{2} < 90^\circ$,so $\tan \frac{A}{2}$ must be positive.
$\tan \frac{A}{2} = \sqrt{\frac{1 - (-3/5)}{1 + (-3/5)}} = \sqrt{\frac{1 + 3/5}{1 - 3/5}} = \sqrt{\frac{8/5}{2/5}} = \sqrt{4} = 2$.

(A) Given that $\tan \frac{\theta}{2} = t$.
We know the standard trigonometric identity for $\cos \theta$ in terms of $\tan \frac{\theta}{2}$ is:
$\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$.
Substituting $t = \tan \frac{\theta}{2}$ into the identity:
$\cos \theta = \frac{1 - t^2}{1 + t^2}$.
Therefore,the expression $\frac{1 - t^2}{1 + t^2}$ is equal to $\cos \theta$.

(C) We evaluate each option:
$(a)$ $\sin 15^\circ = \sin(45^\circ - 30^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4}$,which is irrational.
$(b)$ $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$,which is irrational.
$(c)$ $\sin 15^\circ \cos 15^\circ = \frac{1}{2}(2 \sin 15^\circ \cos 15^\circ) = \frac{1}{2} \sin 30^\circ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$,which is rational.
$(d)$ $\sin 15^\circ \cos 75^\circ = \sin 15^\circ \sin 15^\circ = \sin^2 15^\circ = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \frac{6 + 2 - 2\sqrt{12}}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$,which is irrational.
Thus,the correct option is $(c)$.

(A) We use the half-angle formula: $\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$.
Setting $\theta = 30^\circ$,we get $\cos 15^\circ = \pm \sqrt{\frac{1 + \cos 30^\circ}{2}}$.
Since $15^\circ$ is in the first quadrant,$\cos 15^\circ > 0$.
Therefore,$\cos 15^\circ = \sqrt{\frac{1 + \cos 30^\circ}{2}}$.

(B) Given equation: $2\cos^2 \theta - 2\sin^2 \theta = 1$
Factor out $2$: $2(\cos^2 \theta - \sin^2 \theta) = 1$
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we get:
$2\cos 2\theta = 1$
Divide by $2$: $\cos 2\theta = \frac{1}{2}$
Since $\cos 60^\circ = \frac{1}{2}$,we have:
$2\theta = 60^\circ$
Therefore,$\theta = 30^\circ$.

(C) Given $\sin \alpha = \frac{336}{625}$ and $450^\circ < \alpha < 540^\circ$.
Since $450^\circ < \alpha < 540^\circ$,$\alpha$ lies in the second quadrant (as $450^\circ = 360^\circ + 90^\circ$ and $540^\circ = 360^\circ + 180^\circ$).
In the second quadrant,$\cos \alpha$ is negative.
$\cos \alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - \left( \frac{336}{625} \right)^2} = -\sqrt{\frac{625^2 - 336^2}{625^2}} = -\sqrt{\frac{(625-336)(625+336)}{625^2}} = -\sqrt{\frac{289 \times 961}{625^2}} = -\frac{17 \times 31}{625} = -\frac{527}{625}$.
Now,$450^\circ < \alpha < 540^\circ \implies 225^\circ < \frac{\alpha}{2} < 270^\circ$.
Thus,$\frac{\alpha}{2}$ lies in the third quadrant,where $\cos(\frac{\alpha}{2})$ is negative.
$\cos(\frac{\alpha}{2}) = -\sqrt{\frac{1 + \cos \alpha}{2}} = -\sqrt{\frac{1 - \frac{527}{625}}{2}} = -\sqrt{\frac{98}{1250}} = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.
Finally,$225^\circ < \frac{\alpha}{2} < 270^\circ \implies 112.5^\circ < \frac{\alpha}{4} < 135^\circ$.
Thus,$\frac{\alpha}{4}$ lies in the second quadrant,where $\sin(\frac{\alpha}{4})$ is positive.
$\sin(\frac{\alpha}{4}) = \sqrt{\frac{1 - \cos(\alpha/2)}{2}} = \sqrt{\frac{1 - (-7/25)}{2}} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

(D) Given $\sin x + \cos x = \frac{1}{5}$.
Squaring both sides,we get $(\sin x + \cos x)^2 = (\frac{1}{5})^2$.
$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}{25}$.
Since $\sin^2 x + \cos^2 x = 1$ and $2 \sin x \cos x = \sin 2x$,we have $1 + \sin 2x = \frac{1}{25}$.
$\sin 2x = \frac{1}{25} - 1 = -\frac{24}{25}$.
Now,$\cos^2 2x = 1 - \sin^2 2x = 1 - (-\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$.
$\cos 2x = \pm \frac{7}{25}$.
Since $\sin 2x = -\frac{24}{25}$,$\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{-24/25}{\pm 7/25} = \mp \frac{24}{7}$.
Given the options provided,the correct value is $\frac{24}{7}$.

(C) Given expression: $\cos^2 A(3 - 4\cos^2 A)^2 + \sin^2 A(3 - 4\sin^2 A)^2$
$= (3\cos A - 4\cos^3 A)^2 + (3\sin A - 4\sin^3 A)^2$
Using the triple angle identities $\cos 3A = 4\cos^3 A - 3\cos A$ and $\sin 3A = 3\sin A - 4\sin^3 A$,we have:
$= (-\cos 3A)^2 + (\sin 3A)^2$
$= \cos^2 3A + \sin^2 3A = 1$
Thus,the value of the expression is $1$.