Mix Examples - Introduction to Trigonometry Questions in English

(A) Given,$\cos A = \frac{4}{5}$.
We know that $\sin^2 A + \cos^2 A = 1$,so $\sin A = \sqrt{1 - \cos^2 A}$.
Substituting the value of $\cos A$:
$\sin A = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Now,$\tan A = \frac{\sin A}{\cos A}$.
Substituting the values of $\sin A$ and $\cos A$:
$\tan A = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,the value of $\tan A$ is $\frac{3}{4}$.

(B) Given,$\sin A = \frac{1}{2}$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we find $\cos A$:
$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (\frac{1}{2})^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Now,the value of $\cot A$ is given by:
$\cot A = \frac{\cos A}{\sin A} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$.

(C) Given expression: $E = \operatorname{cosec}(75^{\circ}+\theta) - \sec(15^{\circ}-\theta) - \tan(55^{\circ}+\theta) + \cot(35^{\circ}-\theta)$
Using the complementary angle identities $\operatorname{cosec}(90^{\circ}-A) = \sec A$ and $\cot(90^{\circ}-A) = \tan A$:
$1$. $\operatorname{cosec}(75^{\circ}+\theta) = \operatorname{cosec}(90^{\circ}-(15^{\circ}-\theta)) = \sec(15^{\circ}-\theta)$
$2$. $\cot(35^{\circ}-\theta) = \cot(90^{\circ}-(55^{\circ}+\theta)) = \tan(55^{\circ}+\theta)$
Substituting these into the expression:
$E = \sec(15^{\circ}-\theta) - \sec(15^{\circ}-\theta) - \tan(55^{\circ}+\theta) + \tan(55^{\circ}+\theta)$
$E = 0$
Thus,the value of the expression is $0$.

(D) Given,$\sin \theta = \frac{a}{b}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we can write $\cos \theta = \sqrt{1 - \sin^{2} \theta}$.
Substituting the value of $\sin \theta$:
$\cos \theta = \sqrt{1 - \left(\frac{a}{b}\right)^{2}}$
$\cos \theta = \sqrt{1 - \frac{a^{2}}{b^{2}}}$
$\cos \theta = \sqrt{\frac{b^{2} - a^{2}}{b^{2}}}$
$\cos \theta = \frac{\sqrt{b^{2} - a^{2}}}{b}$.

(A) Given,$\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0\right]$
$\Rightarrow \alpha+\beta=90^{\circ}$
$\Rightarrow \alpha=90^{\circ}-\beta$ ......$(i)$
Now,$\sin (\alpha-\beta)=\sin (90^{\circ}-\beta-\beta)$ [substituting the value from Eq. $(i)$]
$= \sin (90^{\circ}-2 \beta)$
$= \cos 2 \beta$ $\left[\because \sin (90^{\circ}-\theta)=\cos \theta\right]$
Hence,$\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

(B) We have the expression: $P = \tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 89^{\circ}$.
Using the property $\tan(90^{\circ} - \theta) = \cot \theta$,we can write $\tan 89^{\circ} = \cot 1^{\circ}$,$\tan 88^{\circ} = \cot 2^{\circ}$,and so on.
Since $\cot \theta = \frac{1}{\tan \theta}$,we have $\tan 89^{\circ} = \frac{1}{\tan 1^{\circ}}$,$\tan 88^{\circ} = \frac{1}{\tan 2^{\circ}}$,etc.
Substituting these into the expression:
$P = (\tan 1^{\circ} \cdot \cot 1^{\circ}) \cdot (\tan 2^{\circ} \cdot \cot 2^{\circ}) \ldots (\tan 44^{\circ} \cdot \cot 44^{\circ}) \cdot \tan 45^{\circ}$.
Since $\tan \theta \cdot \cot \theta = 1$ and $\tan 45^{\circ} = 1$,the expression becomes:
$P = 1 \cdot 1 \cdot 1 \ldots \cdot 1 = 1$.

(C) Given,$\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ which implies $9 \alpha$ is an acute angle.
Using the trigonometric identity $\cos A = \sin(90^{\circ} - A),$ we can rewrite the equation as:
$\sin(90^{\circ} - 9 \alpha) = \sin \alpha$
Since the sine functions are equal,their angles must be equal:
$90^{\circ} - 9 \alpha = \alpha$
Rearranging the terms to solve for $\alpha$:
$10 \alpha = 90^{\circ}$
$\alpha = 9^{\circ}$
Now,we need to find the value of $\tan 5 \alpha$:
$\tan 5 \alpha = \tan(5 \times 9^{\circ}) = \tan 45^{\circ}$
Since $\tan 45^{\circ} = 1,$ the value is $1$.

(D) We know that in $\triangle ABC$,the sum of the three angles is $180^{\circ}$.
i.e.,$\angle A + \angle B + \angle C = 180^{\circ}$.
Since the triangle is right-angled at $C$,we have $\angle C = 90^{\circ}$ (given).
Substituting this value,we get $\angle A + \angle B + 90^{\circ} = 180^{\circ}$.
Therefore,$A + B = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Now,we need to find the value of $\cos(A + B)$.
Substituting $A + B = 90^{\circ}$,we get $\cos(90^{\circ}) = 0$.

(A) Given,$\sin A + \sin^2 A = 1$.
From this,we can write $\sin A = 1 - \sin^2 A$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we know that $1 - \sin^2 A = \cos^2 A$.
Therefore,$\sin A = \cos^2 A$.
Squaring both sides,we get $\sin^2 A = (\cos^2 A)^2 = \cos^4 A$.
Now,substitute $\sin^2 A = 1 - \cos^2 A$ into the equation:
$1 - \cos^2 A = \cos^4 A$.
Rearranging the terms,we get $\cos^2 A + \cos^4 A = 1$.

(B) Given,$\sin \alpha = \frac{1}{2} = \sin 30^{\circ}$ $\left[\because \sin 30^{\circ} = \frac{1}{2}\right]$
$\Rightarrow \alpha = 30^{\circ}$
And,$\cos \beta = \frac{1}{2} = \cos 60^{\circ}$ $\left[\because \cos 60^{\circ} = \frac{1}{2}\right]$
$\Rightarrow \beta = 60^{\circ}$
Therefore,$\alpha + \beta = 30^{\circ} + 60^{\circ} = 90^{\circ}$

(C) Given expression: $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$
Using the identity $\sin(90^{\circ}-\theta) = \cos \theta$ and $\cos(90^{\circ}-\theta) = \sin \theta$:
Step $1$: Simplify the fraction.
$\sin 68^{\circ} = \sin(90^{\circ}-22^{\circ}) = \cos 22^{\circ}$
$\cos 22^{\circ} = \cos(90^{\circ}-68^{\circ}) = \sin 68^{\circ}$
So,the fraction becomes $\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}} = \frac{1}{1} = 1$.
Step $2$: Simplify the remaining terms.
$\sin 27^{\circ} = \sin(90^{\circ}-63^{\circ}) = \cos 63^{\circ}$
So,$\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ} = \sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ} = \sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ} = 1$.
Step $3$: Add the results.
$1 + 1 = 2$.

(D) Given,$4 \tan \theta = 3$
$\Rightarrow \tan \theta = \frac{3}{4}$ ..........$(i)$
To evaluate the expression $\frac{4 \sin \theta - \cos \theta}{4 \sin \theta + \cos \theta}$,divide both the numerator and the denominator by $\cos \theta$:
$= \frac{\frac{4 \sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}{\frac{4 \sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta}}$
$= \frac{4 \tan \theta - 1}{4 \tan \theta + 1}$
Now,substitute the value of $\tan \theta = \frac{3}{4}$ from equation $(i)$:
$= \frac{4(\frac{3}{4}) - 1}{4(\frac{3}{4}) + 1}$
$= \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$.

(A) Given,$\sin \theta - \cos \theta = 0$.
$\sin \theta = \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta} = 1$.
$\tan \theta = 1$ $[\because \tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\tan 45^{\circ} = 1]$.
$\tan \theta = \tan 45^{\circ}$.
$\theta = 45^{\circ}$.
Now,$\sin^4 \theta + \cos^4 \theta = \sin^4 45^{\circ} + \cos^4 45^{\circ}$.
$= (\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4$ $[\because \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}]$.
$= \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.

(B) We know that $\sin (90^{\circ}-\theta) = \cos \theta$ and $\cos (90^{\circ}-\theta) = \sin \theta$.
Given expression: $\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)$
Using the identity $\sin A = \cos (90^{\circ}-A)$:
$\sin (45^{\circ}+\theta) = \cos (90^{\circ}-(45^{\circ}+\theta))$
$= \cos (90^{\circ}-45^{\circ}-\theta)$
$= \cos (45^{\circ}-\theta)$
Substituting this back into the original expression:
$\cos (45^{\circ}-\theta)-\cos (45^{\circ}-\theta) = 0$

(B) False.
The expression $\sin \theta + \cos \theta$ can be written as $\sqrt{2} (\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta) = \sqrt{2} \sin(\theta + 45^{\circ})$.
For $\theta = 0^{\circ}$,the value is $\sin 0^{\circ} + \cos 0^{\circ} = 0 + 1 = 1$.
Since the value can be equal to $1$ (e.g.,at $\theta = 0^{\circ}$ or $\theta = 90^{\circ}$),the statement that it is 'always greater than $1$' is false.

(TRUE) True
We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
Therefore,$\tan 47^{\circ} = \tan(90^{\circ} - 43^{\circ}) = \cot 43^{\circ}$.
Substituting this into the expression:
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}} = \frac{\cot 43^{\circ}}{\cot 43^{\circ}} = 1$.
Thus,the given statement is True.

(B) False
We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Therefore,$\sin 67^{\circ} = \sin(90^{\circ} - 23^{\circ}) = \cos 23^{\circ}$.
Substituting this into the expression:
$\cos^{2} 23^{\circ} - \sin^{2} 67^{\circ} = \cos^{2} 23^{\circ} - (\cos 23^{\circ})^{2}$
$= \cos^{2} 23^{\circ} - \cos^{2} 23^{\circ}$
$= 0$.
Since $0$ is neither positive nor negative,the statement that the value is positive is False.

(B) The statement is False.
We know that in the interval $0^{\circ} < \theta < 90^{\circ}$,the function $\sin \theta$ is an increasing function,while $\cos \theta$ is a decreasing function.
At $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
For any angle $\theta$ such that $45^{\circ} < \theta < 90^{\circ}$,$\sin \theta > \cos \theta$.
Since $80^{\circ}$ lies in the interval $(45^{\circ}, 90^{\circ})$,we have $\sin 80^{\circ} > \cos 80^{\circ}$.
Therefore,$(\sin 80^{\circ} - \cos 80^{\circ}) > 0$,which means the value is positive,not negative.

(A) True.
Given expression: $\sqrt{(1-\cos^2 \theta) \sec^2 \theta}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,we have $1 - \cos^2 \theta = \sin^2 \theta$.
Substituting this into the expression: $\sqrt{\sin^2 \theta \cdot \sec^2 \theta}$
Since $\sec \theta = \frac{1}{\cos \theta}$,we can write: $\sqrt{\sin^2 \theta \cdot \frac{1}{\cos^2 \theta}} = \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}}$
Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get: $\sqrt{\tan^2 \theta} = \tan \theta$.
Thus,the statement is true.

(A) True.
Given that $\cos A + \cos^2 A = 1$.
Rearranging the terms,we get $\cos A = 1 - \cos^2 A$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we know that $1 - \cos^2 A = \sin^2 A$.
Therefore,$\cos A = \sin^2 A$.
Squaring both sides,we get $\cos^2 A = (\sin^2 A)^2 = \sin^4 A$.
We know that $\cos^2 A = 1 - \sin^2 A$.
Substituting this into the equation $\cos^2 A = \sin^4 A$,we get $1 - \sin^2 A = \sin^4 A$.
Rearranging the terms,we get $\sin^2 A + \sin^4 A = 1$.

(B) False.
$L$.$H$.$S$. $= (\tan \theta + 2)(2 \tan \theta + 1)$
$= 2 \tan^2 \theta + \tan \theta + 4 \tan \theta + 2$
$= 2 \tan^2 \theta + 5 \tan \theta + 2$
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$:
$= 2(\sec^2 \theta - 1) + 5 \tan \theta + 2$
$= 2 \sec^2 \theta - 2 + 5 \tan \theta + 2$
$= 2 \sec^2 \theta + 5 \tan \theta$
Since $2 \sec^2 \theta + 5 \tan \theta \neq 5 \tan \theta + \sec^2 \theta$,the given statement is False.

(B) False.
Given that $a$ is a positive number and $a \neq 1,$ we apply the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality.
For any two positive numbers $a$ and $\frac{1}{a},$ the $AM$ is $\frac{a + \frac{1}{a}}{2}$ and the $GM$ is $\sqrt{a \cdot \frac{1}{a}} = 1.$
Since $AM > GM$ for $a \neq 1,$ we have $\frac{a + \frac{1}{a}}{2} > 1,$ which implies $(a + \frac{1}{a}) > 2.$
If we assume $2 \sin \theta = a + \frac{1}{a},$ then $2 \sin \theta > 2,$ which means $\sin \theta > 1.$
However,we know that the range of $\sin \theta$ is $[-1, 1],$ so $\sin \theta$ can never be greater than $1.$
Therefore,the statement is False.

(B) False.
Given that $a$ and $b$ are two distinct numbers such that $ab > 0$.
We know that for any two distinct positive numbers,the Arithmetic Mean $(AM)$ is strictly greater than the Geometric Mean $(GM)$.
$AM > GM$
$\frac{a^2 + b^2}{2} > \sqrt{a^2 b^2}$
$\frac{a^2 + b^2}{2} > ab$
Dividing both sides by $ab$ (since $ab > 0$):
$\frac{a^2 + b^2}{2ab} > 1$
Since $\cos \theta = \frac{a^2 + b^2}{2ab}$,this implies $\cos \theta > 1$.
However,the range of $\cos \theta$ is $[-1, 1]$,meaning $\cos \theta$ cannot be greater than $1$.
Therefore,the given statement is False.

(N/A) We know the fundamental trigonometric identity: $\sin^{2} \theta + \cos^{2} \theta = 1$.
Taking the cube of both sides,we get:
$(\sin^{2} \theta + \cos^{2} \theta)^{3} = (1)^{3}$
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$,where $a = \sin^{2} \theta$ and $b = \cos^{2} \theta$:
$(\sin^{2} \theta)^{3} + (\cos^{2} \theta)^{3} + 3(\sin^{2} \theta)(\cos^{2} \theta)(\sin^{2} \theta + \cos^{2} \theta) = 1$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,the equation becomes:
$\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta (1) = 1$
Thus,$\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$. Hence proved.

(N/A) $L$.$H$.$S$. $= (\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta$
$= [(\sin^{2} \theta - \cos^{2} \theta)(\sin^{2} \theta + \cos^{2} \theta) + 1] \operatorname{cosec}^{2} \theta$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,we have:
$= (\sin^{2} \theta - \cos^{2} \theta + 1) \operatorname{cosec}^{2} \theta$
Substitute $1 - \cos^{2} \theta = \sin^{2} \theta$:
$= (\sin^{2} \theta + \sin^{2} \theta) \operatorname{cosec}^{2} \theta$
$= (2 \sin^{2} \theta) \operatorname{cosec}^{2} \theta$
$= 2 (\sin^{2} \theta \cdot \operatorname{cosec}^{2} \theta)$
Since $\sin \theta \cdot \operatorname{cosec} \theta = 1$,we get:
$= 2(1) = 2 = \text{R.H.S.}$

(N/A) Given $\alpha + \beta = 90^{\circ}$,we have $\beta = 90^{\circ} - \alpha$.
Substituting $\beta$ in the expression:
$\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sqrt{\cos \alpha \operatorname{cosec}(90^{\circ} - \alpha) - \cos \alpha \sin(90^{\circ} - \alpha)}$
Using trigonometric identities $\operatorname{cosec}(90^{\circ} - \alpha) = \sec \alpha$ and $\sin(90^{\circ} - \alpha) = \cos \alpha$:
$= \sqrt{\cos \alpha \sec \alpha - \cos \alpha \cos \alpha}$
Since $\cos \alpha \sec \alpha = 1$ and $\cos \alpha \cos \alpha = \cos^{2} \alpha$:
$= \sqrt{1 - \cos^{2} \alpha}$
Using the identity $1 - \cos^{2} \alpha = \sin^{2} \alpha$:
$= \sqrt{\sin^{2} \alpha} = \sin \alpha$.
Hence,the expression is proved.

(N/A) Given: $\sin \theta + \cos \theta = \sqrt{3}$.
Squaring both sides:
$(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$1 + 2 \sin \theta \cos \theta = 3$
$2 \sin \theta \cos \theta = 2$
$\sin \theta \cos \theta = 1$.
Now,consider the expression $\tan \theta + \cot \theta$:
$\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$.
Substituting $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin \theta \cos \theta = 1$:
$\tan \theta + \cot \theta = \frac{1}{1} = 1$.
Hence,proved.

(N/A) $L$.$H$.$S$. $= \frac{\sin \theta}{1+\cos \theta} + \frac{1+\cos \theta}{\sin \theta}$
$= \frac{\sin^2 \theta + (1+\cos \theta)^2}{\sin \theta(1+\cos \theta)}$
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$ $\quad [\because (a+b)^2 = a^2 + b^2 + 2ab]$
$= \frac{(\sin^2 \theta + \cos^2 \theta) + 1 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$
$= \frac{1 + 1 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$ $\quad [\because \sin^2 \theta + \cos^2 \theta = 1]$
$= \frac{2 + 2 \cos \theta}{\sin \theta(1+\cos \theta)}$
$= \frac{2(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$
$= \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{R.H.S.}$ $\quad [\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$

(N/A) $L$.$H$.$S$. $= \frac{\tan A}{1+\sec A} + \frac{\tan A}{\sec A-1}$
$= \tan A \left( \frac{\sec A - 1 + 1 + \sec A}{(\sec A + 1)(\sec A - 1)} \right)$
$= \tan A \left( \frac{2 \sec A}{\sec^2 A - 1} \right)$
$= \tan A \left( \frac{2 \sec A}{\tan^2 A} \right) \quad [\because \sec^2 A - 1 = \tan^2 A]$
$= \frac{2 \sec A}{\tan A} = \frac{2 \cdot \frac{1}{\cos A}}{\frac{\sin A}{\cos A}} = \frac{2}{\sin A} = 2 \operatorname{cosec} A = \text{R.H.S.}$

(N/A) Given,$\tan A = \frac{3}{4} = \frac{P}{B} = \frac{\text{Perpendicular}}{\text{Base}}$.
Let $P = 3k$ and $B = 4k$.
By Pythagoras theorem,
$H^2 = P^2 + B^2 = (3k)^2 + (4k)^2$
$H^2 = 9k^2 + 16k^2 = 25k^2$
$\Rightarrow H = 5k$ [since,side length cannot be negative].
Now,$\sin A = \frac{P}{H} = \frac{3k}{5k} = \frac{3}{5}$ and $\cos A = \frac{B}{H} = \frac{4k}{5k} = \frac{4}{5}$.
Therefore,$\sin A \cos A = \frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$.
Hence proved.

$LHS$ $= (\sin \alpha + \cos \alpha)(\tan \alpha + \cot \alpha)$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} \right)$ $\left[ \because \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \text{ and } \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \right]$
$= (\sin \alpha + \cos \alpha) \left( \frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha} \right)$
$= (\sin \alpha + \cos \alpha) \cdot \frac{1}{\sin \alpha \cos \alpha}$ $\left[ \because \sin^2 \alpha + \cos^2 \alpha = 1 \right]$
$= \frac{\sin \alpha}{\sin \alpha \cos \alpha} + \frac{\cos \alpha}{\sin \alpha \cos \alpha}$
$= \frac{1}{\cos \alpha} + \frac{1}{\sin \alpha}$
$= \sec \alpha + \operatorname{cosec} \alpha = \text{RHS}$

(A) We know the trigonometric values: $\tan 60^{\circ} = \sqrt{3}$,$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 30^{\circ} = \sqrt{3}$.
First,evaluate the $R$.$H$.$S$.:
$\text{R.H.S.} = \tan^{3} 60^{\circ} - 2 \sin 60^{\circ} = (\sqrt{3})^{3} - 2 \left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3}$.
Next,evaluate the $L$.$H$.$S$.:
$\text{L.H.S.} = (\sqrt{3} + 1)(3 - \cot 30^{\circ}) = (\sqrt{3} + 1)(3 - \sqrt{3})$.
Factor out $\sqrt{3}$ from the second bracket:
$\text{L.H.S.} = (\sqrt{3} + 1) \sqrt{3}(\sqrt{3} - 1) = \sqrt{3}(\sqrt{3} + 1)(\sqrt{3} - 1)$.
Using the identity $(a+b)(a-b) = a^2 - b^2$:
$\text{L.H.S.} = \sqrt{3}((\sqrt{3})^2 - 1^2) = \sqrt{3}(3 - 1) = \sqrt{3}(2) = 2\sqrt{3}$.
Since $\text{L.H.S.} = \text{R.H.S.}$,the identity is proved.

(N/A) $L$.$H$.$S$. $= 1 + \frac{\cot^{2} \alpha}{1 + \operatorname{cosec} \alpha}$
$= 1 + \frac{\cos^{2} \alpha / \sin^{2} \alpha}{1 + 1 / \sin \alpha} \quad [\because \cot \theta = \frac{\cos \theta}{\sin \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= 1 + \frac{\cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)} = \frac{\sin \alpha(1 + \sin \alpha) + \cos^{2} \alpha}{\sin \alpha(1 + \sin \alpha)}$
$= \frac{\sin \alpha + (\sin^{2} \alpha + \cos^{2} \alpha)}{\sin \alpha(1 + \sin \alpha)} \quad [\because \sin^{2} \theta + \cos^{2} \theta = 1]$
$= \frac{\sin \alpha + 1}{\sin \alpha(1 + \sin \alpha)} = \frac{1}{\sin \alpha} \quad [\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}]$
$= \operatorname{cosec} \alpha = \text{R.H.S.}$

(N/A) $L$.$H$.$S$. $= \tan \theta + \tan (90^{\circ} - \theta)$
Since $\tan (90^{\circ} - \theta) = \cot \theta$,we have:
$= \tan \theta + \cot \theta$
$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$
$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$= \frac{1}{\sin \theta \cos \theta}$
$= \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}$
$= \sec \theta \operatorname{cosec} \theta$
Since $\operatorname{cosec} \theta = \sec (90^{\circ} - \theta)$,we get:
$= \sec \theta \sec (90^{\circ} - \theta) = \text{R.H.S.}$

(C) Given that,$\sqrt{3} \tan \theta = 1$.
$\tan \theta = \frac{1}{\sqrt{3}}$.
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.
Now,substitute $\theta = 30^{\circ}$ into the expression $\sin^{2} \theta - \cos^{2} \theta$:
$\sin^{2} 30^{\circ} - \cos^{2} 30^{\circ} = \left(\frac{1}{2}\right)^{2} - \left(\frac{\sqrt{3}}{2}\right)^{2}$.
$= \frac{1}{4} - \frac{3}{4}$.
$= \frac{1 - 3}{4} = -\frac{2}{4} = -\frac{1}{2}$.

(A) Given expression: $(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)$
Using the algebraic identity $(a-b)(a+b) = a^{2}-b^{2}$,we have $(1-\sin \theta)(1+\sin \theta) = 1-\sin ^{2} \theta$.
So,the expression becomes $(1+\tan ^{2} \theta)(1-\sin ^{2} \theta)$.
We know that $1+\tan ^{2} \theta = \sec ^{2} \theta$ and $1-\sin ^{2} \theta = \cos ^{2} \theta$.
Substituting these,we get $\sec ^{2} \theta \cdot \cos ^{2} \theta$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we have $\sec ^{2} \theta = \frac{1}{\cos ^{2} \theta}$.
Therefore,$\frac{1}{\cos ^{2} \theta} \cdot \cos ^{2} \theta = 1$.

(B) Given equation: $2 \sin^{2} \theta - \cos^{2} \theta = 2$
Using the identity $\cos^{2} \theta = 1 - \sin^{2} \theta$,we substitute it into the equation:
$2 \sin^{2} \theta - (1 - \sin^{2} \theta) = 2$
Simplify the expression:
$2 \sin^{2} \theta - 1 + \sin^{2} \theta = 2$
$3 \sin^{2} \theta - 1 = 2$
Add $1$ to both sides:
$3 \sin^{2} \theta = 3$
Divide by $3$:
$\sin^{2} \theta = 1$
Taking the square root:
$\sin \theta = 1$ (considering the principal value)
Since $\sin 90^{\circ} = 1$,we have:
$\theta = 90^{\circ}$

(N/A) $L$.$H$.$S$. $= \frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \cdot \tan \left(30^{\circ}-\theta\right)}$
Using the identity $\cos \theta = \sin(90^{\circ}-\theta)$ and $\tan \theta = \cot(90^{\circ}-\theta)$:
Numerator: $\cos ^{2}\left(45^{\circ}+\theta\right) + \sin ^{2}\left(90^{\circ}-(45^{\circ}-\theta)\right) = \cos ^{2}\left(45^{\circ}+\theta\right) + \sin ^{2}\left(45^{\circ}+\theta\right) = 1$
Denominator: $\tan \left(60^{\circ}+\theta\right) \cdot \cot \left(90^{\circ}-(30^{\circ}-\theta)\right) = \tan \left(60^{\circ}+\theta\right) \cdot \cot \left(60^{\circ}+\theta\right) = \tan \left(60^{\circ}+\theta\right) \cdot \frac{1}{\tan \left(60^{\circ}+\theta\right)} = 1$
Therefore,$\frac{1}{1} = 1 = \text{R.H.S.}$

(N/A) $L$.$H$.$S$. $= \tan ^{4} \theta + \tan ^{2} \theta$
$= \tan ^{2} \theta (\tan ^{2} \theta + 1)$
$= \tan ^{2} \theta \cdot \sec ^{2} \theta$ (Since $\sec ^{2} \theta = \tan ^{2} \theta + 1$)
$= (\sec ^{2} \theta - 1) \cdot \sec ^{2} \theta$ (Since $\tan ^{2} \theta = \sec ^{2} \theta - 1$)
$= \sec ^{4} \theta - \sec ^{2} \theta = \text{R.H.S.}$

(N/A) Given,$\operatorname{cosec} \theta + \cot \theta = p$.
We know that $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
Substituting these values,we get $\frac{1 + \cos \theta}{\sin \theta} = p$.
Squaring both sides,we get $\frac{(1 + \cos \theta)^{2}}{\sin^{2} \theta} = p^{2}$.
Using the identity $\sin^{2} \theta = 1 - \cos^{2} \theta = (1 - \cos \theta)(1 + \cos \theta)$,we have:
$\frac{(1 + \cos \theta)^{2}}{(1 - \cos \theta)(1 + \cos \theta)} = p^{2} \Rightarrow \frac{1 + \cos \theta}{1 - \cos \theta} = p^{2}$.
Applying the componendo and dividendo rule:
$\frac{p^{2} - 1}{p^{2} + 1} = \frac{(1 + \cos \theta) - (1 - \cos \theta)}{(1 + \cos \theta) + (1 - \cos \theta)}$.
Simplifying the expression:
$\frac{p^{2} - 1}{p^{2} + 1} = \frac{1 + \cos \theta - 1 + \cos \theta}{1 + \cos \theta + 1 - \cos \theta} = \frac{2 \cos \theta}{2} = \cos \theta$.
Thus,$\cos \theta = \frac{p^{2} - 1}{p^{2} + 1}$. Hence proved.

(A) $L.H.S. = \sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}$
$= \sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}$ $\left[\because \sec \theta = \frac{1}{\cos \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}\right]$
$= \sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}} = \sqrt{\frac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta = 1\right]$
$= \frac{1}{\sin \theta \cdot \cos \theta} = \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$ $\left[\because 1 = \sin ^{2} \theta+\cos ^{2} \theta\right]$
$= \frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta} + \frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$
$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$ $\left[\because \tan \theta = \frac{\sin \theta}{\cos \theta} \text{ and } \cot \theta = \frac{\cos \theta}{\sin \theta}\right]$
$= \tan \theta + \cot \theta = R.H.S.$

(A) Given equation: $1+\sin ^{2} \theta=3 \sin \theta \cos \theta$
Divide both sides by $\cos^{2} \theta$:
$\frac{1}{\cos^{2} \theta} + \frac{\sin^{2} \theta}{\cos^{2} \theta} = \frac{3 \sin \theta \cos \theta}{\cos^{2} \theta}$
Using the identities $\sec^{2} \theta = \frac{1}{\cos^{2} \theta}$,$\tan \theta = \frac{\sin \theta}{\cos \theta}$,and $\sec^{2} \theta = 1 + \tan^{2} \theta$:
$(1 + \tan^{2} \theta) + \tan^{2} \theta = 3 \tan \theta$
Rearranging the terms to form a quadratic equation:
$2 \tan^{2} \theta - 3 \tan \theta + 1 = 0$
Factorizing the quadratic equation:
$2 \tan^{2} \theta - 2 \tan \theta - \tan \theta + 1 = 0$
$2 \tan \theta (\tan \theta - 1) - 1 (\tan \theta - 1) = 0$
$(2 \tan \theta - 1) (\tan \theta - 1) = 0$
Therefore,$\tan \theta - 1 = 0$ or $2 \tan \theta - 1 = 0$
$\tan \theta = 1$ or $\tan \theta = \frac{1}{2}$.
Hence proved.

(N/A) Given,$\sin \theta + 2 \cos \theta = 1$.
Squaring both sides,we get:
$(\sin \theta + 2 \cos \theta)^2 = 1^2$
$\sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 1$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$ and $\cos^2 \theta = 1 - \sin^2 \theta$:
$(1 - \cos^2 \theta) + 4(1 - \sin^2 \theta) + 4 \sin \theta \cos \theta = 1$
$1 - \cos^2 \theta + 4 - 4 \sin^2 \theta + 4 \sin \theta \cos \theta = 1$
$5 - (\cos^2 \theta + 4 \sin^2 \theta - 4 \sin \theta \cos \theta) = 1$
Rearranging the terms:
$4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = 5 - 1$
$4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = 4$
Recognizing the algebraic identity $(a - b)^2 = a^2 + b^2 - 2ab$,where $a = 2 \sin \theta$ and $b = \cos \theta$:
$(2 \sin \theta - \cos \theta)^2 = 4$
Taking the square root on both sides:
$2 \sin \theta - \cos \theta = 2$
Hence proved.

(A) Given: $\tan \theta + \sec \theta = l$ .....$(i)$
We know the identity: $\sec^{2} \theta - \tan^{2} \theta = 1$.
This can be written as $(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1$.
Substituting the value of $(\sec \theta + \tan \theta)$ from equation $(i)$:
$(\sec \theta - \tan \theta) \cdot l = 1$
$\sec \theta - \tan \theta = \frac{1}{l}$ .....$(ii)$
Now,adding equation $(i)$ and equation $(ii)$:
$(\tan \theta + \sec \theta) + (\sec \theta - \tan \theta) = l + \frac{1}{l}$
$2 \sec \theta = \frac{l^{2} + 1}{l}$
Dividing both sides by $2$:
$\sec \theta = \frac{l^{2} + 1}{2l}$.
Hence,it is proved.

(N/A) Given that,$\sin \theta + \cos \theta = p$ ......$(i)$
And $\sec \theta + \operatorname{cosec} \theta = q$
$\Rightarrow \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q$ $\left[\because \sec \theta = \frac{1}{\cos \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}\right]$
$\Rightarrow \frac{\sin \theta + \cos \theta}{\sin \theta \cdot \cos \theta} = q$
$\Rightarrow \frac{p}{\sin \theta \cdot \cos \theta} = q$ [from Eq. $(i)$]
$\Rightarrow \sin \theta \cdot \cos \theta = \frac{p}{q}$ ......$(ii)$
Now,squaring Eq. $(i)$ on both sides:
$(\sin \theta + \cos \theta)^2 = p^2$
$\Rightarrow (\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cdot \cos \theta = p^2$ $\left[\because (a+b)^2 = a^2 + 2ab + b^2\right]$
$\Rightarrow 1 + 2 \sin \theta \cdot \cos \theta = p^2$ $\left[\because \sin^2 \theta + \cos^2 \theta = 1\right]$
Substitute $\sin \theta \cdot \cos \theta = \frac{p}{q}$ from Eq. $(ii)$:
$\Rightarrow 1 + 2 \left(\frac{p}{q}\right) = p^2$
$\Rightarrow 1 + \frac{2p}{q} = p^2$
Multiply both sides by $q$:
$\Rightarrow q + 2p = p^2 q$
$\Rightarrow 2p = p^2 q - q$
$\Rightarrow 2p = q(p^2 - 1)$
Hence,$q(p^2 - 1) = 2p$ is proved.

(N/A) Given: $a \sin \theta + b \cos \theta = c$ $(1)$
Let $x = a \cos \theta - b \sin \theta$ $(2)$
Squaring both equations $(1)$ and $(2)$ and adding them:
$(a \sin \theta + b \cos \theta)^2 + (a \cos \theta - b \sin \theta)^2 = c^2 + x^2$
$a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta = c^2 + x^2$
$a^2(\sin^2 \theta + \cos^2 \theta) + b^2(\cos^2 \theta + \sin^2 \theta) = c^2 + x^2$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$a^2(1) + b^2(1) = c^2 + x^2$
$a^2 + b^2 = c^2 + x^2$
$x^2 = a^2 + b^2 - c^2$
$x = \pm \sqrt{a^2 + b^2 - c^2}$
Therefore,$a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2}$.

(A) $L$.$H$.$S$. $= \frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}$
$= \frac{1+\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}{1+\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}$ $\left[\because \sec \theta=\frac{1}{\cos \theta} \text{ and } \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$= \frac{\cos \theta+1-\sin \theta}{\cos \theta+1+\sin \theta} = \frac{(\cos \theta+1)-\sin \theta}{(\cos \theta+1)+\sin \theta}$
Using the identity $1 = \sec^2 \theta - \tan^2 \theta$,we can write:
$= \frac{(\sec^2 \theta - \tan^2 \theta) + (\sec \theta - \tan \theta)}{1 + \sec \theta + \tan \theta}$
$= \frac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) + (\sec \theta - \tan \theta)}{1 + \sec \theta + \tan \theta}$
$= \frac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta + 1)}{1 + \sec \theta + \tan \theta}$
$= \sec \theta - \tan \theta$
$= \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1-\sin \theta}{\cos \theta} = \text{R.H.S.}$
Hence proved.

(B) In $\Delta ABC$, $AC = 5$ and hypotenuse $BC = 13$, with $m \angle A = 90^\circ$.
By the Pythagorean theorem:
$AB^2 = BC^2 - AC^2$
$AB^2 = 13^2 - 5^2 = 169 - 25 = 144$
$AB = \sqrt{144} = 12$
Now, $\tan B = \frac{\text{side opposite to } \angle B}{\text{side adjacent to } \angle B} = \frac{AC}{AB} = \frac{5}{12}$.

(C) Given: $\tan A = \frac{1}{\sqrt{3}}$.
In a right-angled triangle,$\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AC}$.
Therefore,$\frac{BC}{AC} = \frac{1}{\sqrt{3}}$.
Let $BC = k$ and $AC = \sqrt{3}k$,where $k > 0$.
Using the Pythagorean theorem: $AB^2 = BC^2 + AC^2$.
$AB^2 = k^2 + (\sqrt{3}k)^2 = k^2 + 3k^2 = 4k^2$.
Thus,$AB = 2k$.
Now,$\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{k}{2k} = \frac{1}{2}$.
Alternatively,since $\tan A = \frac{1}{\sqrt{3}}$,we know that $A = 30^{\circ}$.
Therefore,$\sin A = \sin 30^{\circ} = \frac{1}{2}$.

(D) Given that in $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\cos B = \frac{1}{2}$.
We know that $\cos B = \frac{\text{adjacent side to } \angle B}{\text{hypotenuse}} = \frac{BC}{AB}$.
Therefore,$\frac{BC}{AB} = \frac{1}{2}$. Let $BC = k$ and $AB = 2k$,where $k > 0$.
Using the Pythagorean theorem,$AC^2 + BC^2 = AB^2$.
$AC^2 + k^2 = (2k)^2 = 4k^2$.
$AC^2 = 3k^2 \implies AC = k\sqrt{3}$.
Now,$\operatorname{cosec} A = \frac{\text{hypotenuse}}{\text{side opposite to } \angle A} = \frac{AB}{BC}$.
$\operatorname{cosec} A = \frac{2k}{k} = 2$.