Trigonometric Identities Questions in English

(N/A) In $\triangle ACB$,we have:
$AC = \sqrt{AB^2 - BC^2} = \sqrt{(29)^2 - (21)^2}$
$= \sqrt{(29 - 21)(29 + 21)} = \sqrt{(8)(50)} = \sqrt{400} = 20$ units.
So,$\sin \theta = \frac{AC}{AB} = \frac{20}{29}$ and $\cos \theta = \frac{BC}{AB} = \frac{21}{29}$.
Now,
$(i)$ $\cos^2 \theta + \sin^2 \theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2 = \frac{441 + 400}{841} = \frac{841}{841} = 1$.
$(ii)$ $\cos^2 \theta - \sin^2 \theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2 = \frac{441 - 400}{841} = \frac{41}{841}$.

(A) Given equation: $\sin 3A = \cos(A - 26^{\circ})$
We know that $\sin \theta = \cos(90^{\circ} - \theta)$.
Therefore,$\sin 3A = \cos(90^{\circ} - 3A)$.
Substituting this into the given equation:
$\cos(90^{\circ} - 3A) = \cos(A - 26^{\circ})$
Since the angles are acute,we can equate them:
$90^{\circ} - 3A = A - 26^{\circ}$
Rearranging the terms to solve for $A$:
$90^{\circ} + 26^{\circ} = A + 3A$
$116^{\circ} = 4A$
$A = \frac{116^{\circ}}{4} = 29^{\circ}$
Thus,the value of $A$ is $29^{\circ}$.

(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 $0$.

(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:
$P = 1 \cdot 1 \cdot 1 \ldots \cdot 1 = 1$.

(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 $\cos^2 A + \cos^4 A = 1$.

(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.
$1 + 1 = 2$.

(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.

(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.

(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.

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

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

(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) $(i)$ We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = \tan(90^{\circ} - 42^{\circ}) \tan(90^{\circ} - 67^{\circ}) \tan 42^{\circ} \tan 67^{\circ}$
$= \cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= (\cot 42^{\circ} \tan 42^{\circ}) (\cot 67^{\circ} \tan 67^{\circ})$
$= (1)(1) = 1$.
$(ii)$ We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
$\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = \cos(90^{\circ} - 52^{\circ}) \cos(90^{\circ} - 38^{\circ}) - \sin 38^{\circ} \sin 52^{\circ}$
$= \sin 52^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$.

(N/A) Given that,$\tan A = \cot B$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.

(B) Given that,$\sec 4A = \operatorname{cosec}(A - 20^{\circ})$.
We know that $\sec \theta = \operatorname{cosec}(90^{\circ} - \theta)$.
Therefore,$\operatorname{cosec}(90^{\circ} - 4A) = \operatorname{cosec}(A - 20^{\circ})$.
Equating the angles,we get $90^{\circ} - 4A = A - 20^{\circ}$.
Rearranging the terms,$90^{\circ} + 20^{\circ} = A + 4A$.
$110^{\circ} = 5A$.
$A = \frac{110^{\circ}}{5} = 22^{\circ}$.
Thus,the value of $A$ is $22^{\circ}$.

(N/A) We know that for a triangle $ABC$,the sum of interior angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle B + \angle C = 180^{\circ} - \angle A$
Dividing both sides by $2$,we get:
$\frac{\angle B + \angle C}{2} = \frac{180^{\circ} - \angle A}{2} = 90^{\circ} - \frac{\angle A}{2}$
Taking the sine on both sides:
$\sin \left(\frac{B+C}{2}\right) = \sin \left(90^{\circ} - \frac{A}{2}\right)$
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we have:
$\sin \left(\frac{B+C}{2}\right) = \cos \left(\frac{A}{2}\right)$
Hence,it is proved.

(N/A) We know the fundamental trigonometric identity: $\cos^2 A + \sin^2 A = 1$.
$1$. To express $\cos A$ in terms of $\sin A$:
$\cos^2 A = 1 - \sin^2 A$
$\cos A = \sqrt{1 - \sin^2 A}$ (taking the positive root for acute angle $A$)
$2$. To express $\tan A$ in terms of $\sin A$:
$\tan A = \frac{\sin A}{\cos A}$
Substituting the value of $\cos A$:
$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$
$3$. To express $\sec A$ in terms of $\sin A$:
$\sec A = \frac{1}{\cos A}$
Substituting the value of $\cos A$:
$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$

(A) $LHS = \sec A(1-\sin A)(\sec A+\tan A)$
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.

(N/A) We start with the Left Hand Side $(LHS)$:
$LHS = \frac{\cot A - \cos A}{\cot A + \cos A}$
Substitute $\cot A = \frac{\cos A}{\sin A}$:
$LHS = \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A}$
Factor out $\cos A$ from the numerator and the denominator:
$LHS = \frac{\cos A \left( \frac{1}{\sin A} - 1 \right)}{\cos A \left( \frac{1}{\sin A} + 1 \right)}$
Cancel $\cos A$ from both:
$LHS = \frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1}$
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we substitute it:
$LHS = \frac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1} = RHS$
Hence,the identity is proved.

(A) To prove the identity,we divide the numerator and the denominator of the $LHS$ by $\cos \theta$ to express it in terms of $\tan \theta$ and $\sec \theta.$
$LHS = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$
Rearranging the terms,we get:
$LHS = \frac{(\tan \theta + \sec \theta) - 1}{(\tan \theta - \sec \theta) + 1}$
Using the identity $\sec^2 \theta - \tan^2 \theta = 1,$ we substitute $1$ in the numerator:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{(\tan \theta - \sec \theta) + 1}$
Factorizing the numerator using $a^2 - b^2 = (a - b)(a + b)$:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{(\tan \theta - \sec \theta) + 1}$
Taking $(\tan \theta + \sec \theta)$ as a common factor:
$LHS = \frac{(\tan \theta + \sec \theta) [1 - (\sec \theta - \tan \theta)]}{(\tan \theta - \sec \theta) + 1}$
$LHS = \frac{(\tan \theta + \sec \theta) [1 - \sec \theta + \tan \theta]}{(\tan \theta - \sec \theta + 1)}$
Canceling the common term $(\tan \theta - \sec \theta + 1)$:
$LHS = \tan \theta + \sec \theta$
To get the $RHS,$ multiply and divide by $(\sec \theta - \tan \theta)$:
$LHS = \frac{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)}{\sec \theta - \tan \theta} = \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = RHS.$

(N/A) We know that,$\operatorname{cosec}^{2} A = 1 + \cot^{2} A$.
Since $\sin A = \frac{1}{\operatorname{cosec} A}$,we have $\sin^{2} A = \frac{1}{\operatorname{cosec}^{2} A} = \frac{1}{1 + \cot^{2} A}$.
Taking the square root,$\sin A = \frac{1}{\sqrt{1 + \cot^{2} A}}$.
Next,we know that $\tan A = \frac{1}{\cot A}$.
Finally,using the identity $\sec^{2} A = 1 + \tan^{2} A$,we substitute $\tan A = \frac{1}{\cot A}$:
$\sec^{2} A = 1 + \left(\frac{1}{\cot A}\right)^{2} = 1 + \frac{1}{\cot^{2} A} = \frac{\cot^{2} A + 1}{\cot^{2} A}$.
Taking the square root,$\sec A = \frac{\sqrt{1 + \cot^{2} A}}{\cot A}$.

We know that,$\cos A = \frac{1}{\sec A}$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we have $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}$.
Thus,$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$.
Using the identity $\tan^2 A + 1 = \sec^2 A$,we have $\tan^2 A = \sec^2 A - 1$.
Thus,$\tan A = \sqrt{\sec^2 A - 1}$.
Since $\cot A = \frac{1}{\tan A}$,we have $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$.
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$.

(D) The given expression is $\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$.
We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
Substituting $\cos 65^{\circ} = \cos(90^{\circ} - 25^{\circ}) = \sin 25^{\circ}$ and $\sin 65^{\circ} = \sin(90^{\circ} - 25^{\circ}) = \cos 25^{\circ}$:
$= (\sin 25^{\circ})(\sin 25^{\circ}) + (\cos 25^{\circ})(\cos 25^{\circ})$
$= \sin^{2} 25^{\circ} + \cos^{2} 25^{\circ}$
Using the identity $\sin^{2} A + \cos^{2} A = 1$,we get:
$= 1$

(A) Given expression: $9 \sec^{2} A - 9 \tan^{2} A$
Factor out the common term $9$:
$= 9(\sec^{2} A - \tan^{2} A)$
Using the trigonometric identity $\sec^{2} A - \tan^{2} A = 1$:
$= 9(1)$
$= 9$

(B) Given expression: $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
Convert the trigonometric ratios into $\sin \theta$ and $\cos \theta$:
$= \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)$
Simplify the terms inside the brackets:
$= \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$
Apply the algebraic identity $(a+b)(a-b) = a^2 - b^2$,where $a = (\sin \theta + \cos \theta)$ and $b = 1$:
$= \frac{(\sin \theta + \cos \theta)^2 - (1)^2}{\sin \theta \cos \theta}$
Expand the numerator using $(a+b)^2 = a^2 + b^2 + 2ab$:
$= \frac{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$= \frac{1 + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta}$
$= \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} = 2$

(C) Given expression: $(\sec A + \tan A)(1 - \sin A)$
Step $1$: Convert $\sec A$ and $\tan A$ into terms of $\sin A$ and $\cos A$:
$(\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A)$
Step $2$: Combine the fractions:
$(\frac{1 + \sin A}{\cos A})(1 - \sin A)$
Step $3$: Multiply the numerators:
$\frac{(1 + \sin A)(1 - \sin A)}{\cos A}$
Step $4$: Use the identity $(a + b)(a - b) = a^2 - b^2$:
$\frac{1 - \sin^2 A}{\cos A}$
Step $5$: Use the identity $1 - \sin^2 A = \cos^2 A$:
$\frac{\cos^2 A}{\cos A} = \cos A$

(D) We know the trigonometric identities: $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$.
Substituting these into the expression:
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A} = \frac{\sec^2 A}{\csc^2 A}$
Since $\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$,we have:
$= \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

(A) To prove: $(\operatorname{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
Consider the $L.H.S. = (\operatorname{cosec} \theta - \cot \theta)^2$
Using the trigonometric identities $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we get:
$L.H.S. = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2$
$= \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$
Since $a^2 - b^2 = (a - b)(a + b)$,we can write $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$:
$= \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}$
$= \frac{1 - \cos \theta}{1 + \cos \theta} = R.H.S.$
Hence,$L.H.S. = R.H.S.$

(N/A) To prove: $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
$L.H.S. = \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}$
Taking the common denominator:
$= \frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)(\cos A)}$
Expanding the numerator:
$= \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{(1+\sin A)(\cos A)}$
Using the identity $\sin^2 A + \cos^2 A = 1$:
$= \frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{(1+\sin A)(\cos A)}$
$= \frac{1 + 1 + 2\sin A}{(1+\sin A)(\cos A)}$
$= \frac{2 + 2\sin A}{(1+\sin A)(\cos A)}$
Factoring out $2$:
$= \frac{2(1+\sin A)}{(1+\sin A)(\cos A)}$
$= \frac{2}{\cos A} = 2 \sec A$
$= R.H.S.$

(A) $L.H.S. = \frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$= \frac{1}{(\sin \theta-\cos \theta)} \left[ \frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{\sin^3 \theta-\cos^3 \theta}{\sin \theta \cos \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta \cos \theta)}{\sin \theta \cos \theta} \right]$
$= \frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \operatorname{cosec} \theta \sec \theta + 1 = R.H.S.$

(A) To prove: $\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
Step $1$: Simplify the Left Hand Side $(L.H.S.)$.
$L.H.S. = \frac{1+\sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$
Step $2$: Simplify the fraction.
$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1$
Step $3$: Simplify the Right Hand Side $(R.H.S.)$ to match the $L.H.S.$ or multiply the $L.H.S.$ by $\frac{1-\cos A}{1-\cos A}$.
$= (1 + \cos A) \times \frac{1-\cos A}{1-\cos A}$
Step $4$: Use the identity $\sin^2 A + \cos^2 A = 1$,so $1 - \cos^2 A = \sin^2 A$.
$= \frac{1 - \cos^2 A}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A}$
Thus,$L.H.S. = R.H.S.$

(N/A) To prove: $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$
$L.H.S. = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$
Divide the numerator and the denominator by $\sin A$:
$= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$
$= \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A}$
Using the identity $1 = \operatorname{cosec}^{2} A - \cot^{2} A$ in the numerator:
$= \frac{\cot A + \operatorname{cosec} A - (\operatorname{cosec}^{2} A - \cot^{2} A)}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) - (\operatorname{cosec} A - \cot A)(\operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Factor out $(\cot A + \operatorname{cosec} A)$:
$= \frac{(\cot A + \operatorname{cosec} A) [1 - (\operatorname{cosec} A - \cot A)]}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) (1 - \operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Since $(1 - \operatorname{cosec} A + \cot A) = (\cot A - \operatorname{cosec} A + 1)$,they cancel out:
$= \cot A + \operatorname{cosec} A = R.H.S.$

(N/A) To prove the identity $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$:
$L.H.S. = \sqrt{\frac{1+\sin A}{1-\sin A}}$
Multiply the numerator and denominator inside the square root by $(1 + \sin A)$:
$= \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
Using the identity $(1 - \sin^2 A) = \cos^2 A$:
$= \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$
Taking the square root:
$= \frac{1+\sin A}{\cos A}$
$= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$
$= R.H.S.$

(A) Given: $L.H.S. = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$
Step $1$: Factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$
Step $2$: Use the identity $\sin^2 \theta + \cos^2 \theta = 1$,which implies $\cos^2 \theta = 1 - \sin^2 \theta$.
Substitute this into the denominator:
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta [2(1 - \sin^2 \theta) - 1]}$
Step $3$: Simplify the expression inside the bracket in the denominator.
$2(1 - \sin^2 \theta) - 1 = 2 - 2 \sin^2 \theta - 1 = 1 - 2 \sin^2 \theta$
Step $4$: Substitute back into the expression.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (1 - 2 \sin^2 \theta)}$
Step $5$: Cancel the common term $(1 - 2 \sin^2 \theta)$.
$L.H.S. = \frac{\sin \theta}{\cos \theta} = \tan \theta = R.H.S.$
Hence,the identity is proved.

(N/A) To prove: $(\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
$L.H.S. = (\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$:
$= (\sin^2 A + \operatorname{cosec}^2 A + 2 \sin A \operatorname{cosec} A) + (\cos^2 A + \sec^2 A + 2 \cos A \sec A)$
$= (\sin^2 A + \cos^2 A) + \operatorname{cosec}^2 A + \sec^2 A + 2 \sin A \left(\frac{1}{\sin A}\right) + 2 \cos A \left(\frac{1}{\cos A}\right)$
Since $\sin^2 A + \cos^2 A = 1$,$\operatorname{cosec}^2 A = 1 + \cot^2 A$,and $\sec^2 A = 1 + \tan^2 A$:
$= 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 2(1) + 2(1)$
$= 1 + 1 + \cot^2 A + 1 + \tan^2 A + 2 + 2$
$= 7 + \tan^2 A + \cot^2 A$
$= R.H.S.$

(A) To prove: $(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
$L.H.S. = (\operatorname{cosec} A - \sin A)(\sec A - \cos A)$
$= (\frac{1}{\sin A} - \sin A)(\frac{1}{\cos A} - \cos A)$
$= (\frac{1 - \sin^2 A}{\sin A})(\frac{1 - \cos^2 A}{\cos A})$
$= (\frac{\cos^2 A}{\sin A})(\frac{\sin^2 A}{\cos A})$
$= \sin A \cos A$
$R.H.S. = \frac{1}{\tan A + \cot A}$
$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$
$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
$= \frac{\sin A \cos A}{1} = \sin A \cos A$
Since $L.H.S. = R.H.S.$,the identity is proved.

(A) First,consider the expression $\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$.
Using the identities $1+\tan ^{2} A = \sec ^{2} A$ and $1+\cot ^{2} A = \operatorname{cosec}^{2} A$,we get:
$\frac{\sec ^{2} A}{\operatorname{cosec}^{2} A} = \frac{1/\cos ^{2} A}{1/\sin ^{2} A} = \frac{\sin ^{2} A}{\cos ^{2} A} = \tan ^{2} A$.
Next,consider the expression $\left(\frac{1-\tan A}{1-\cot A}\right)^{2}$.
Substitute $\cot A = \frac{1}{\tan A}$:
$\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^{2} = \left(\frac{1-\tan A}{\frac{\tan A - 1}{\tan A}}\right)^{2} = \left(\frac{(1-\tan A) \cdot \tan A}{-(1-\tan A)}\right)^{2} = (-\tan A)^{2} = \tan ^{2} A$.
Since both parts equal $\tan ^{2} A$,the identity is proved.

(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:
$= \sec \theta \sec (90^{\circ} - \theta) = \text{R.H.S.}$

(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:
$\theta = 90^{\circ}$