Trigonometry Questions in English

(C) The formula for the angle $\theta$ subtended by an arc of length $l$ at the centre of a circle of radius $r$ is given by $\theta = \frac{l}{r}$,where $\theta$ is in radians.
Given,radius $(r) = 3 \text{ m}$ and arc length $(l) = 1 \text{ m}$.
Substituting these values into the formula:
$\theta = \frac{1}{3} \text{ radian}$.
Therefore,the correct option is $C$.

(B) The circumference of the circular wire is the length of the wire.
Given radius $r_1 = 7\,cm$.
Length of the wire $L = 2\pi r_1 = 2 \times \pi \times 7 = 14\pi\,cm$.
This wire is now bent into an arc of a circle with radius $r_2 = 12\,cm$.
The length of the arc $s = L = 14\pi\,cm$.
The formula for the angle $\theta$ subtended by an arc at the center is $\theta = \frac{s}{r_2}$ (in radians).
$\theta = \frac{14\pi}{12} = \frac{7\pi}{6}$ radians.
To convert radians to degrees,multiply by $\frac{180^o}{\pi}$.
$\theta = \frac{7\pi}{6} \times \frac{180^o}{\pi} = 7 \times 30^o = 210^o$.

(B) The relationship between the arc length $(l)$,the radius $(r)$,and the angle subtended at the centre in radians $( heta)$ is given by the formula: $l = r \theta$.
Given:
Arc length $(l)$ = $15 \ cm$
Angle $( heta)$ = $3/4 \ \text{radian}$
Substituting the values into the formula:
$15 = r \times (3/4)$
$r = 15 \times (4/3)$
$r = 5 \times 4$
$r = 20 \ cm$.
Therefore,the radius of the circle is $20 \ cm$.

(D) Given the equation $\cos \theta = x + \frac{1}{x}$.
Multiplying by $x$,we get $x^2 - x \cos \theta + 1 = 0$.
Since $x$ is a real value,the discriminant $D$ of this quadratic equation must be greater than or equal to $0$.
$D = b^2 - 4ac = (-\cos \theta)^2 - 4(1)(1) \ge 0$.
$\cos^2 \theta - 4 \ge 0 \Rightarrow \cos^2 \theta \ge 4$.
However,the range of $\cos \theta$ is $[-1, 1]$,so $\cos^2 \theta$ must be in the range $[0, 1]$.
Since $\cos^2 \theta \ge 4$ is impossible for any real $\theta$,there is no real value of $x$ that satisfies the given equation for any $\theta$.
Therefore,no value of $\theta$ is possible.

(B) We are given the equation $(a + b)^2 = 4ab \sin^2 \theta$.
Since the range of the sine function is $[-1, 1]$,the value of $\sin^2 \theta$ must satisfy $0 \le \sin^2 \theta \le 1$.
Therefore,we have $\sin^2 \theta = \frac{(a + b)^2}{4ab} \le 1$.
This implies $(a + b)^2 \le 4ab$.
Rearranging the terms,we get $(a + b)^2 - 4ab \le 0$.
Expanding the square,we get $a^2 + 2ab + b^2 - 4ab \le 0$,which simplifies to $a^2 - 2ab + b^2 \le 0$.
This is equivalent to $(a - b)^2 \le 0$.
Since the square of any real number cannot be negative,the only possibility is $(a - b)^2 = 0$,which implies $a = b$.

(A) We know that for any real angle $\theta$,$\cos^2 \theta \le 1$,which implies $\sec^2 \theta \ge 1$.
Given the equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$,we must have $\frac{4xy}{(x + y)^2} \ge 1$.
Since $(x + y)^2$ is always non-negative,we can multiply both sides by $(x + y)^2$ to get $4xy \ge (x + y)^2$.
Expanding the right side,we get $4xy \ge x^2 + 2xy + y^2$.
Rearranging the terms,we get $0 \ge x^2 - 2xy + y^2$,which simplifies to $(x - y)^2 \le 0$.
Since the square of any real number cannot be negative,the only possibility is $(x - y)^2 = 0$,which means $x = y$.

(A) The function $f(x) = \tan x$ is strictly increasing in the interval $(0, \pi/2)$.
Since $1 \text{ radian} \approx 57.3^\circ$ and $2 \text{ radians} \approx 114.6^\circ$,we observe that $1 \text{ radian}$ lies in the first quadrant $(0, \pi/2)$ where $\tan x$ is positive and increasing.
However,$2 \text{ radians}$ lies in the second quadrant $(\pi/2, \pi)$ where $\tan x$ is negative.
Since any positive value is greater than any negative value,$\tan 1 > \tan 2$ is correct.

(B) The correct relation is $\sin 1 > \sin 1^\circ $.
In trigonometry,the value of $1$ radian is approximately $57.3^\circ$.
Since the function $f(x) = \sin x$ is an increasing function in the interval $[0, \pi/2]$,and $57.3^\circ > 1^\circ$,it follows that $\sin(57.3^\circ) > \sin(1^\circ)$.
Therefore,$\sin 1 > \sin 1^\circ$.

(A) We know that $\tan(90^\circ - \theta) = \cot \theta$.
Therefore,$\tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ$.
Similarly,$\tan 88^\circ = \cot 2^\circ$,and so on.
The expression can be written as: $(\tan 1^\circ \cdot \tan 89^\circ) \cdot (\tan 2^\circ \cdot \tan 88^\circ) \dots (\tan 44^\circ \cdot \tan 46^\circ) \cdot \tan 45^\circ$.
Since $\tan \theta \cdot \cot \theta = 1$,each pair $(\tan \theta \cdot \tan(90^\circ - \theta)) = 1$.
There are $44$ such pairs,and $\tan 45^\circ = 1$.
Thus,the product is $1 \times 1 \times \dots \times 1 = 1$.

(D) Given,$\sin \theta + \text{cosec} \theta = 2$.
Since $\text{cosec} \theta = \frac{1}{\sin \theta}$,the equation becomes $\sin \theta + \frac{1}{\sin \theta} = 2$.
Multiplying by $\sin \theta$,we get $\sin^2 \theta + 1 = 2 \sin \theta$.
Rearranging gives $\sin^2 \theta - 2 \sin \theta + 1 = 0$,which is $(\sin \theta - 1)^2 = 0$.
Therefore,$\sin \theta = 1$.
Substituting this value into the expression: $\sin^{10} \theta + \text{cosec}^{10} \theta = (1)^{10} + \left(\frac{1}{1}\right)^{10} = 1 + 1 = 2$.

(C) Given that $\sin \theta + \csc \theta = 2$.
Squaring both sides of the equation:
$(\sin \theta + \csc \theta)^2 = 2^2$
$\sin^2 \theta + \csc^2 \theta + 2 \sin \theta \csc \theta = 4$
Since $\csc \theta = \frac{1}{\sin \theta}$,we have $\sin \theta \csc \theta = 1$.
Substituting this value into the equation:
$\sin^2 \theta + \csc^2 \theta + 2(1) = 4$
$\sin^2 \theta + \csc^2 \theta + 2 = 4$
$\sin^2 \theta + \csc^2 \theta = 4 - 2 = 2$.

(C) Given that $\sin \theta + \cos \theta = m$ and $\sec \theta + \csc \theta = n$.
We need to find the value of $n(m + 1)(m - 1)$,which is $n(m^2 - 1)$.
First,square the first equation: $m^2 = (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1 + 2 \sin \theta \cos \theta$.
Thus,$m^2 - 1 = 2 \sin \theta \cos \theta$.
Now,substitute $n = \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{m}{\sin \theta \cos \theta}$.
Therefore,$n(m^2 - 1) = \left( \frac{m}{\sin \theta \cos \theta} \right) (2 \sin \theta \cos \theta) = 2m$.

(A) Given: $\sin \theta + \cos \theta = 1$
Squaring both sides of the equation,we get:
$(\sin \theta + \cos \theta)^2 = 1^2$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$:
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 1$
We know the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:
$1 + 2 \sin \theta \cos \theta = 1$
Subtracting $1$ from both sides:
$2 \sin \theta \cos \theta = 0$
Dividing by $2$:
$\sin \theta \cos \theta = 0$

(C) Given $\sin \theta = \frac{24}{25}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ and $\tan \theta$ are negative.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\cos^2 \theta = 1 - (\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$.
Since $\theta$ is in the second quadrant,$\cos \theta = -\sqrt{\frac{49}{625}} = -\frac{7}{25}$.
Then,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{24/25}{-7/25} = -\frac{24}{7}$.
Also,$\sec \theta = \frac{1}{\cos \theta} = -\frac{25}{7}$.
Therefore,$\sec \theta + \tan \theta = -\frac{25}{7} + (-\frac{24}{7}) = -\frac{49}{7} = -7$.

(C) We are given that $\csc A + \cot A = \frac{11}{2}$.
We know the identity $\csc^2 A - \cot^2 A = 1$,which can be factored as $(\csc A - \cot A)(\csc A + \cot A) = 1$.
Substituting the given value: $(\csc A - \cot A) \times \frac{11}{2} = 1$.
Thus,$\csc A - \cot A = \frac{2}{11}$.
Now,subtract the two equations:
$(\csc A + \cot A) - (\csc A - \cot A) = \frac{11}{2} - \frac{2}{11}$.
$2 \cot A = \frac{121 - 4}{22} = \frac{117}{22}$.
$\cot A = \frac{117}{44}$.
Since $\tan A = \frac{1}{\cot A}$,we have $\tan A = \frac{44}{117}$.

(C) Given $5 \tan \theta = 4$,which implies $\tan \theta = \frac{4}{5}$.
To evaluate the expression $\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta}$,divide both the numerator and the denominator by $\cos \theta$:
$= \frac{\frac{5 \sin \theta}{\cos \theta} - \frac{3 \cos \theta}{\cos \theta}}{\frac{5 \sin \theta}{\cos \theta} + \frac{2 \cos \theta}{\cos \theta}}$
$= \frac{5 \tan \theta - 3}{5 \tan \theta + 2}$
Substitute $\tan \theta = \frac{4}{5}$ into the expression:
$= \frac{5(\frac{4}{5}) - 3}{5(\frac{4}{5}) + 2}$
$= \frac{4 - 3}{4 + 2}$
$= \frac{1}{6}$.

(C) Given that $\sin \theta = - \frac{1}{\sqrt{2}}$ and $\tan \theta = 1$.
$1$. The value of $\sin \theta$ is negative,which implies that $\theta$ must lie in either the third or fourth quadrant.
$2$. The value of $\tan \theta$ is positive,which implies that $\theta$ must lie in either the first or third quadrant.
$3$. For both conditions to be satisfied simultaneously,$\theta$ must lie in the third quadrant,where both $\sin \theta$ is negative and $\tan \theta$ is positive.
Therefore,$\theta$ lies in the third quadrant.

(B) Given $\sin \theta = -\frac{4}{5}$ and $\theta$ lies in the $III$ quadrant $(180^\circ < \theta < 270^\circ)$.
Since $\theta$ is in the $III$ quadrant,$\cos \theta$ is 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 need to find $\cos \frac{\theta}{2}$. Using the half-angle formula: $\cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}$.
Since $180^\circ < \theta < 270^\circ$,dividing by $2$ gives $90^\circ < \frac{\theta}{2} < 135^\circ$.
This means $\frac{\theta}{2}$ lies in the $II$ quadrant,where cosine is negative.
Therefore,$\cos \frac{\theta}{2} = -\sqrt{\frac{1 + (-3/5)}{2}} = -\sqrt{\frac{2/5}{2}} = -\sqrt{\frac{1}{5}} = -\frac{1}{\sqrt{5}}$.

(A) Given,$\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.
Also,$\cos (\alpha + \beta ) = \frac{1}{2}$. Since $\cos 60^\circ = \frac{1}{2}$,we have $\alpha + \beta = 60^\circ$ $(ii)$.
Adding equations $(i)$ and $(ii)$:
$(\alpha - \beta ) + (\alpha + \beta ) = 30^\circ + 60^\circ$
$2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ$.
Substituting $\alpha = 45^\circ$ in equation $(ii)$:
$45^\circ + \beta = 60^\circ \Rightarrow \beta = 15^\circ$.
Thus,$\alpha = 45^\circ$ and $\beta = 15^\circ$ satisfy both conditions.

(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: $((m + 2)\tan \theta + (2m - 1))^2 = (2m + 1)^2 \sec^2 \theta$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$ and letting $t = \tan \theta$:
$(m + 2)^2 t^2 + 2(m + 2)(2m - 1)t + (2m - 1)^2 = (2m + 1)^2 (1 + t^2)$.
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:
$(m^2 + 4m + 4 - 4m^2 - 4m - 1)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1 - 4m^2 - 4m - 1) = 0$.
$(-3m^2 + 3)t^2 + (4m^2 + 6m - 4)t - 8m = 0$.
$3(1 - m^2)t^2 + (4m^2 + 6m - 4)t - 8m = 0$.
Factoring the quadratic equation:
$(3t - 4)((1 - m^2)t + 2m) = 0$.
Thus,$t = \frac{4}{3}$ or $t = \frac{2m}{m^2 - 1}$.

(D) Given $3\tan A + 4 = 0$,we have $\tan A = -\frac{4}{3}$.
Since $A$ lies in the second quadrant,$\sin A$ is positive and $\cos A$ is negative.
Using the identity $\sec^2 A = 1 + \tan^2 A$,we get $\sec^2 A = 1 + (-\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$.
Thus,$\sec A = -\frac{5}{3}$ (since $\cos A$ is negative in the second quadrant),which implies $\cos A = -\frac{3}{5}$.
Also,$\sin A = \tan A \times \cos A = (-\frac{4}{3}) \times (-\frac{3}{5}) = \frac{4}{5}$.
Finally,$\cot A = \frac{1}{\tan A} = -\frac{3}{4}$.
Substituting these values into the expression $2\cot A - 5\cos A + \sin A$:
$= 2(-\frac{3}{4}) - 5(-\frac{3}{5}) + \frac{4}{5}$
$= -\frac{3}{2} + 3 + \frac{4}{5}$
$= \frac{-15 + 30 + 8}{10} = \frac{23}{10}$.

(B) Given: $\sin x + \sin y = 3(\cos y - \cos x)$
Rearranging the terms: $\sin x + 3\cos x = 3\cos y - \sin y$ ... $(i)$
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 $\sin \alpha = \frac{3}{\sqrt{10}}$,then $\tan \alpha = 3$.
The equation becomes: $\sin x \cos \alpha + \cos x \sin \alpha = \cos y \sin \alpha - \sin y \cos \alpha$
$\sin(x + \alpha) = \sin(\alpha - y)$
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 condition for $G.P.$ is $(\cos A)^2 = \sin A \cdot \tan A$.
Substituting $\tan A = \frac{\sin A}{\cos A}$,we get $\cos^2 A = \sin A \cdot \frac{\sin A}{\cos A}$.
Multiplying both sides by $\cos A$,we get $\cos^3 A = \sin^2 A$.
We know the identity $\sin^2 A = 1 - \cos^2 A$.
Substituting this into our equation,we get $\cos^3 A = 1 - \cos^2 A$.
Rearranging the terms,we get $\cos^3 A + \cos^2 A = 1$.

(B) Let the expression be $E = \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} + \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}}$.
Simplifying the expression by taking the common denominator:
$E = \frac{\sqrt{1 - \sin \theta} \cdot \sqrt{1 - \sin \theta} + \sqrt{1 + \sin \theta} \cdot \sqrt{1 + \sin \theta}}{\sqrt{(1 + \sin \theta)(1 - \sin \theta)}}$
$E = \frac{(1 - \sin \theta) + (1 + \sin \theta)}{\sqrt{1 - \sin^2 \theta}} = \frac{2}{\sqrt{\cos^2 \theta}} = \frac{2}{|\cos \theta|}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ is negative.
Therefore,$|\cos \theta| = -\cos \theta$.
Thus,$E = \frac{2}{-\cos \theta} = -2 \sec \theta$.

(D) Given expression: $\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }}$
Substitute $\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$ and $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}$:
$= \frac{{\sin \theta }}{{1 - \frac{{\cos \theta }}{{\sin \theta }}}} + \frac{{\cos \theta }}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}}$
$= \frac{{\sin \theta }}{{\frac{{\sin \theta - \cos \theta }}{{\sin \theta }}}} + \frac{{\cos \theta }}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}}$
$= \frac{{{{\sin }^2}\theta }}{{\sin \theta - \cos \theta }} + \frac{{{{\cos }^2}\theta }}{{\cos \theta - \sin \theta }}$
$= \frac{{{{\sin }^2}\theta }}{{\sin \theta - \cos \theta }} - \frac{{{{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}$
$= \frac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{\sin \theta - \cos \theta }}$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$= \frac{{(\sin \theta - \cos \theta )(\sin \theta + \cos \theta )}}{{\sin \theta - \cos \theta }}$
$= \sin \theta + \cos \theta $.

(B) We are given that $\tan \theta + \sec \theta = e^x$ --- $(i)$
We know the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$.
This can be factored as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value from $(i)$,we get $(\sec \theta - \tan \theta) e^x = 1$,which implies $\sec \theta - \tan \theta = e^{-x}$ --- $(ii)$
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 $\sec \theta = \frac{e^x + e^{-x}}{2}$.
Therefore,$\cos \theta = \frac{2}{e^x + e^{-x}}$.

(A) Given: $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$
Rearranging the terms,we get: $\cos \theta = \sqrt{2} \sin \theta + \sin \theta$
$\cos \theta = (\sqrt{2} + 1) \sin \theta$
To isolate $\sin \theta$,multiply both sides by $(\sqrt{2} - 1)$:
$(\sqrt{2} - 1) \cos \theta = (\sqrt{2} - 1)(\sqrt{2} + 1) \sin \theta$
$(\sqrt{2} - 1) \cos \theta = (2 - 1) \sin \theta$
$\sqrt{2} \cos \theta - \cos \theta = \sin \theta$
Rearranging to find $\cos \theta + \sin \theta$:
$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$

(B) We are given that $\sec \theta + \tan \theta = p$ ........... $(i)$
We know the identity $\sec^2 \theta - \tan^2 \theta = 1$,which can be factored as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value from $(i)$,we get $(\sec \theta - \tan \theta) \cdot p = 1$,which implies $\sec \theta - \tan \theta = \frac{1}{p}$ ........... $(ii)$
To find $\tan \theta$,subtract equation $(ii)$ from equation $(i)$:
$(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = p - \frac{1}{p}$
$2 \tan \theta = \frac{p^2 - 1}{p}$
$\tan \theta = \frac{p^2 - 1}{2p}$.

(B) Given that $x = \sec \theta + \tan \theta$.
We know the identity $\sec^2 \theta - \tan^2 \theta = 1$,which can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Therefore,$\frac{1}{x} = \frac{1}{\sec \theta + \tan \theta} = \sec \theta - \tan \theta$.
Now,calculating $x + \frac{1}{x}$:
$x + \frac{1}{x} = (\sec \theta + \tan \theta) + (\sec \theta - \tan \theta)$.
$x + \frac{1}{x} = 2 \sec \theta$.

(D) Given $x + \frac{1}{x} = 2\cos \alpha$.
Multiplying by $x$, we get $x^2 - (2\cos \alpha)x + 1 = 0$.
Using the quadratic formula, $x = \frac{2\cos \alpha \pm \sqrt{4\cos^2 \alpha - 4}}{2} = \cos \alpha \pm i\sin \alpha$.
Let $x = e^{i\alpha} = \cos \alpha + i\sin \alpha$. Then $\frac{1}{x} = e^{-i\alpha} = \cos \alpha - i\sin \alpha$.
By De Moivre's Theorem, $x^n = e^{in\alpha} = \cos n\alpha + i\sin n\alpha$ and $\frac{1}{x^n} = e^{-in\alpha} = \cos n\alpha - i\sin n\alpha$.
Adding these, we get ${x^n} + \frac{1}{{{x^n}}} = (\cos n\alpha + i\sin n\alpha) + (\cos n\alpha - i\sin n\alpha) = 2\cos n\alpha$.

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

(D) Let $S = \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \dots \cdot \tan 89^\circ$.
Using the property $\tan \theta \cdot \tan(90^\circ - \theta) = 1$,we can pair the terms:
$(\tan 1^\circ \cdot \tan 89^\circ) \cdot (\tan 2^\circ \cdot \tan 88^\circ) \cdot \dots \cdot (\tan 44^\circ \cdot \tan 46^\circ) \cdot \tan 45^\circ$.
Each pair equals $1$,and $\tan 45^\circ = 1$. Thus,$S = 1$.
The expression is $e^{\log_{10}(S)} = e^{\log_{10}(1)} = e^0 = 1$.

(C) We know that $\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$.
Substituting these into the expression: $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}$.
Taking the common denominator: $\frac{\cos^2 x - \sin^2 x}{\sin x \cos x}$.
Using the identity $\cos^2 x - \sin^2 x = \cos 2x$ and multiplying the numerator and denominator by $2$: $\frac{2 \cos 2x}{2 \sin x \cos x}$.
Since $2 \sin x \cos x = \sin 2x$,the expression becomes $\frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x$.

(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 $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$:
Numerator: $(1 - \cos A) + \sin A = 2 \sin^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2} = 2 \sin \frac{A}{2} (\sin \frac{A}{2} + \cos \frac{A}{2})$
Denominator: $(1 + \cos A) + \sin A = 2 \cos^2 \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2} = 2 \cos \frac{A}{2} (\cos \frac{A}{2} + \sin \frac{A}{2})$
Dividing the numerator by the denominator:
$\frac{2 \sin \frac{A}{2} (\sin \frac{A}{2} + \cos \frac{A}{2})}{2 \cos \frac{A}{2} (\cos \frac{A}{2} + \sin \frac{A}{2})} = \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} = \tan \frac{A}{2}$
Thus,the correct option is $(c)$.

(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 $2\sin \theta$ from the numerator:
$E = \frac{2\sin \theta [\tan \theta (1 - \tan \theta ) + \sec^2 \theta]}{(1 + \tan \theta )^2}$
Expand the terms inside the bracket:
$E = \frac{2\sin \theta [\tan \theta - \tan^2 \theta + \sec^2 \theta]}{(1 + \tan \theta )^2}$
Use the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$,which implies $\sec^2 \theta - \tan^2 \theta = 1$:
$E = \frac{2\sin \theta [\tan \theta + 1]}{(1 + \tan \theta )^2}$
Cancel the common term $(1 + \tan \theta)$:
$E = \frac{2\sin \theta}{1 + \tan \theta}$.

(D) Let the expression be $E = 1 - \frac{\sin^2 y}{1 + \cos y} + \frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y}$.
First,simplify the first two terms: $1 - \frac{\sin^2 y}{1 + \cos y} = \frac{1 + \cos y - (1 - \cos^2 y)}{1 + \cos y} = \frac{\cos y + \cos^2 y}{1 + \cos y} = \frac{\cos y(1 + \cos y)}{1 + \cos y} = \cos y$.
Now,simplify the last two terms: $\frac{1 + \cos y}{\sin y} - \frac{\sin y}{1 - \cos y} = \frac{(1 + \cos y)(1 - \cos y) - \sin^2 y}{\sin y(1 - \cos y)} = \frac{1 - \cos^2 y - \sin^2 y}{\sin y(1 - \cos y)} = \frac{\sin^2 y - \sin^2 y}{\sin y(1 - \cos y)} = 0$.
Adding these 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).
Then $x = k \cos \theta$ and $2y = k \sin \theta$,which implies $y = \frac{k}{2} \sin \theta$.
Substitute these into $(ii)$:
$2(k \cos \theta) \frac{1}{\cos \theta} - (\frac{k}{2} \sin \theta) \frac{1}{\sin \theta} = 3$
$2k - \frac{k}{2} = 3$
$\frac{3k}{2} = 3 \implies k = 2$.
Thus,$x = 2 \cos \theta$ and $y = \sin \theta$.
Now,$x^2 + 4y^2 = (2 \cos \theta)^2 + 4(\sin \theta)^2$
$= 4 \cos^2 \theta + 4 \sin^2 \theta$
$= 4(\cos^2 \theta + \sin^2 \theta) = 4(1) = 4$.

(D) Given: $\tan A + \cot A = 4$
Squaring both sides:
$(\tan A + \cot A)^2 = 4^2$
$\tan^2 A + \cot^2 A + 2 \tan A \cot A = 16$
Since $\tan A \cot A = 1$,we have:
$\tan^2 A + \cot^2 A + 2(1) = 16$
$\tan^2 A + \cot^2 A = 14$
Now,squaring both sides again:
$(\tan^2 A + \cot^2 A)^2 = 14^2$
$\tan^4 A + \cot^4 A + 2 \tan^2 A \cot^2 A = 196$
$\tan^4 A + \cot^4 A + 2(1)^2 = 196$
$\tan^4 A + \cot^4 A = 196 - 2 = 194$
Thus,the correct option is $D$.

(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 $x = \frac{1 - \sin \phi}{\cos \phi}$. Then $\frac{1}{x} = \frac{\cos \phi}{1 - \sin \phi} = \frac{\cos \phi(1 + \sin \phi)}{1 - \sin^2 \phi} = \frac{\cos \phi(1 + \sin \phi)}{\cos^2 \phi} = \frac{1 + \sin \phi}{\cos \phi}$.
Similarly,$y = \frac{1 + \cos \phi}{\sin \phi}$. Then $\frac{1}{y} = \frac{\sin \phi}{1 + \cos \phi} = \frac{\sin \phi(1 - \cos \phi)}{1 - \cos^2 \phi} = \frac{\sin \phi(1 - \cos \phi)}{\sin^2 \phi} = \frac{1 - \cos \phi}{\sin \phi}$.
Alternatively,using the identity $\sec \phi - \tan \phi = \tan(\frac{\pi}{4} - \frac{\phi}{2})$ and $\csc \phi + \cot \phi = \cot(\frac{\phi}{2})$,we can relate $x$ and $y$.
Let's test the relation $x = \frac{y - 1}{y + 1}$.
Substituting $y = \frac{1 + \cos \phi}{\sin \phi}$ into $\frac{y - 1}{y + 1}$:
$\frac{\frac{1 + \cos \phi}{\sin \phi} - 1}{\frac{1 + \cos \phi}{\sin \phi} + 1} = \frac{1 + \cos \phi - \sin \phi}{1 + \cos \phi + \sin \phi} = \frac{2\cos^2(\phi/2) - 2\sin(\phi/2)\cos(\phi/2)}{2\cos^2(\phi/2) + 2\sin(\phi/2)\cos(\phi/2)} = \frac{\cos(\phi/2) - \sin(\phi/2)}{\cos(\phi/2) + \sin(\phi/2)} = \tan(\frac{\pi}{4} - \frac{\phi}{2}) = x$.
Thus,$x = \frac{y - 1}{y + 1}$ is correct.

(B) Given $\tan \theta = \frac{x \sin \phi}{1 - x \cos \phi}$.
Rearranging,we get $x \sin \phi = \tan \theta - x \cos \phi \tan \theta$.
$x(\sin \phi + \cos \phi \tan \theta) = \tan \theta$.
$x \left( \sin \phi + \cos \phi \frac{\sin \theta}{\cos \theta} \right) = \frac{\sin \theta}{\cos \theta}$.
$x \left( \frac{\sin \phi \cos \theta + \cos \phi \sin \theta}{\cos \theta} \right) = \frac{\sin \theta}{\cos \theta}$.
$x \sin(\theta + \phi) = \sin \theta \Rightarrow x = \frac{\sin \theta}{\sin(\theta + \phi)}$.
Similarly,for $\tan \phi = \frac{y \sin \theta}{1 - y \cos \theta}$,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 $p + q = \frac{2 \sin \theta}{1 + \sin \theta + \cos \theta} + \frac{\cos \theta}{1 + \sin \theta}$.
To simplify $p$,multiply the numerator and denominator by $(1 + \sin \theta - \cos \theta)$:
$p = \frac{2 \sin \theta (1 + \sin \theta - \cos \theta)}{(1 + \sin \theta)^2 - \cos^2 \theta} = \frac{2 \sin \theta (1 + \sin \theta - \cos \theta)}{1 + 2 \sin \theta + \sin^2 \theta - \cos^2 \theta}$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$:
$p = \frac{2 \sin \theta (1 + \sin \theta - \cos \theta)}{1 + 2 \sin \theta + \sin^2 \theta - (1 - \sin^2 \theta)} = \frac{2 \sin \theta (1 + \sin \theta - \cos \theta)}{2 \sin \theta + 2 \sin^2 \theta} = \frac{2 \sin \theta (1 + \sin \theta - \cos \theta)}{2 \sin \theta (1 + \sin \theta)} = \frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta} = 1 - \frac{\cos \theta}{1 + \sin \theta}$.
Since $q = \frac{\cos \theta}{1 + \sin \theta}$,we have $p = 1 - q$.
Therefore,$p + q = 1$.

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

(A) Given that $\tan \theta = \frac{a}{b}$.
Since $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{a}{b}$,we can write $\sin \theta = \pm \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos \theta = \pm \frac{b}{\sqrt{a^2 + b^2}}$.
Now,consider the expression $\frac{\sin \theta}{\cos^8 \theta} + \frac{\cos \theta}{\sin^8 \theta}$.
Substituting the values of $\sin \theta$ and $\cos \theta$:
$= \frac{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} \right)}{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} \right)^8} + \frac{\left( \pm \frac{b}{\sqrt{a^2 + b^2}} \right)}{\left( \pm \frac{a}{\sqrt{a^2 + b^2}} \right)^8}$
$= \frac{\pm a (a^2 + b^2)^4}{b^8 \sqrt{a^2 + b^2}} + \frac{\pm b (a^2 + b^2)^4}{a^8 \sqrt{a^2 + b^2}}$
$= \pm \frac{(a^2 + b^2)^4}{\sqrt{a^2 + b^2}} \left( \frac{a}{b^8} + \frac{b}{a^8} \right)$.

(C) Given that $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n.$
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 equations are $x = a \cos^3 \theta$ and $y = b \sin^3 \theta$.
Step $1$: Express $\cos \theta$ and $\sin \theta$ in terms of $x$ and $y$.
$\frac{x}{a} = \cos^3 \theta \implies (\frac{x}{a})^{1/3} = \cos \theta$
$\frac{y}{b} = \sin^3 \theta \implies (\frac{y}{b})^{1/3} = \sin \theta$
Step $2$: Use the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$.
Substituting the expressions from Step $1$:
$((\frac{x}{a})^{1/3})^2 + ((\frac{y}{b})^{1/3})^2 = 1$
$(\frac{x}{a})^{2/3} + (\frac{y}{b})^{2/3} = 1$.
Thus,the correct option is $C$.

(A) Given: $\cot \theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n$.
$1$. From $\cot \theta + \tan \theta = m$:
$\frac{1}{\tan \theta} + \tan \theta = m \implies \frac{1 + \tan^2 \theta}{\tan \theta} = m \implies \sec^2 \theta = m \tan \theta$ --- $(i)$
$2$. From $\sec \theta - \cos \theta = n$:
$\sec \theta - \frac{1}{\sec \theta} = n \implies \frac{\sec^2 \theta - 1}{\sec \theta} = n \implies \tan^2 \theta = n \sec \theta$ --- $(ii)$
$3$. Substitute $\sec^2 \theta = m \tan \theta$ into $(ii)$:
$(\tan^2 \theta)^2 = n^2 \sec^2 \theta \implies \tan^4 \theta = n^2 (m \tan \theta) \implies \tan^3 \theta = n^2 m \implies \tan \theta = (n^2 m)^{1/3}$.
$4$. Similarly,find $\sec \theta$:
Since $\sec^2 \theta = m \tan \theta = m(n^2 m)^{1/3} = (m^4 n^2)^{1/3}$,then $\sec \theta = (m^2 n)^{1/3}$.
$5$. Using the identity $\sec^2 \theta - \tan^2 \theta = 1$:
$(\sec \theta)^2 - (\tan \theta)^2 = 1$
$(m^2 n)^{2/3} - (n^2 m)^{2/3} = 1$ is not the form,let's use $\sec^2 \theta = m \tan \theta$ and $\tan^2 \theta = n \sec \theta$:
$m \tan \theta - n \sec \theta = 1$ is not correct,rather substitute $\tan \theta = (n^2 m)^{1/3}$ and $\sec \theta = (m^2 n)^{1/3}$ into $\sec^2 \theta - \tan^2 \theta = 1$ is not the path.
Actually,$m \tan \theta - n \sec \theta$ is not $1$.
Using $\sec^2 \theta = m \tan \theta$ and $\tan^2 \theta = n \sec \theta$,we have $\sec^2 \theta - \tan^2 \theta = 1 \implies (m \tan \theta) - (n \sec \theta)$ is not $1$.
Wait,$\sec^2 \theta = m \tan \theta$ and $\tan^2 \theta = n \sec \theta$.
Then $\sec^2 \theta = m(n^2 m)^{1/3} = m^{4/3} n^{2/3}$ and $\tan^2 \theta = n(m^2 n)^{1/3} = n^{4/3} m^{2/3}$.
Thus,$m(mn^2)^{1/3} - n(nm^2)^{1/3} = 1$.

(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 substituting $\theta = 0^\circ$,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 $(\sin^2 \theta + \cos^2 \theta)^3 = 1^3$.
Expanding this,we get $\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,this simplifies to $\sin^6 \theta + \cos^6 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta$.
Next,consider $(\sin^2 \theta + \cos^2 \theta)^2 = 1^2$.
Expanding this,we get $\sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta = 1$.
This simplifies to $\sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Now,substitute these into the given expression:
$2(1 - 3 \sin^2 \theta \cos^2 \theta) - 3(1 - 2 \sin^2 \theta \cos^2 \theta) + 1$
$= 2 - 6 \sin^2 \theta \cos^2 \theta - 3 + 6 \sin^2 \theta \cos^2 \theta + 1$
$= (2 - 3 + 1) + (-6 \sin^2 \theta \cos^2 \theta + 6 \sin^2 \theta \cos^2 \theta)$
$= 0 + 0 = 0$.

(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 $\cos^2 x = \sin x$ into the expression:
$E = (\cos^2 x)^6 + 3(\cos^2 x)^5 + 3(\cos^2 x)^4 + (\cos^2 x)^3 - 2$
$E = (\sin x)^6 + 3(\sin x)^5 + 3(\sin x)^4 + (\sin x)^3 - 2$
$E = (\sin^2 x)^3 + 3(\sin^2 x)^2(\sin x) + 3(\sin^2 x)(\sin x)^2 + (\sin x)^3 - 2$
This is in the form of $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,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:
$E = (1)^3 - 2 = 1 - 2 = -1$.

(A) Given the equation: $\cos x + \cos^2 x = 1$.
Rearranging the terms,we get: $\cos x = 1 - \cos^2 x$.
Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$,we know that $1 - \cos^2 x = \sin^2 x$.
Therefore,$\cos x = \sin^2 x$.
Now,we need to find the value of $\sin^2 x + \sin^4 x$.
Substituting $\sin^2 x = \cos x$ into the expression,we get:
$\sin^2 x + (\sin^2 x)^2 = \cos x + (\cos x)^2$.
Since we are given $\cos x + \cos^2 x = 1$,the value of the expression is $1$.