Trigonometric Identities Questions in English

(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: $a \sin \theta + b \cos \theta = c$ $(1)$
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.

(A) In trigonometry,the trigonometric ratios of complementary angles are defined as follows:
$\sin (90^\circ - \theta) = \cos \theta$
$\cos (90^\circ - \theta) = \sin \theta$
$\tan (90^\circ - \theta) = \cot \theta$
$\cot (90^\circ - \theta) = \tan \theta$
$\sec (90^\circ - \theta) = \operatorname{cosec} \theta$
$\operatorname{cosec} (90^\circ - \theta) = \sec \theta$
Therefore,based on the standard trigonometric identity for complementary angles,$\sin (90^\circ - \theta) = \cos \theta$.

(B) In trigonometry,the complementary angle identities state that the trigonometric functions of an angle $(90^\circ - \theta)$ are related to the co-functions of $\theta$.
Specifically,for the secant function,the identity is $\sec (90^\circ - \theta) = \operatorname{cosec} \theta$.
Therefore,the correct option is $B$.

(B) In trigonometry,the trigonometric ratios of complementary angles are defined as follows:
$\sin (90^\circ - \theta) = \cos \theta$
$\cos (90^\circ - \theta) = \sin \theta$
$\tan (90^\circ - \theta) = \cot \theta$
$\cot (90^\circ - \theta) = \tan \theta$
$\sec (90^\circ - \theta) = \operatorname{cosec} \theta$
$\operatorname{cosec} (90^\circ - \theta) = \sec \theta$
Therefore,$\tan (90^\circ - \theta) = \cot \theta$.

(A) Given that $\sin \theta + \sin^2 \theta = 1$.
We can rewrite this as $\sin \theta = 1 - \sin^2 \theta$.
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we know that $1 - \sin^2 \theta = \cos^2 \theta$.
Therefore,$\sin \theta = \cos^2 \theta$.
Now,we need to find the value of $\cos^2 \theta + \cos^4 \theta$.
Substitute $\cos^2 \theta = \sin \theta$ into the expression:
$\cos^2 \theta + \cos^4 \theta = \sin \theta + (\cos^2 \theta)^2$.
Since $\cos^2 \theta = \sin \theta$,then $(\cos^2 \theta)^2 = \sin^2 \theta$.
So,the expression becomes $\sin \theta + \sin^2 \theta$.
Given that $\sin \theta + \sin^2 \theta = 1$,the final value is $1$.

(D) We know that $\sec \theta = \frac{1}{\cos \theta}$ and $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$.
Also,$\sin(90^{\circ} - \theta) = \cos \theta$ and $\cos(90^{\circ} - \theta) = \sin \theta$.
Given expression: $\sec 55^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec} 55^{\circ}$.
Convert $55^{\circ}$ to $(90^{\circ} - 35^{\circ})$:
$= \sec(90^{\circ} - 35^{\circ}) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \operatorname{cosec}(90^{\circ} - 35^{\circ})$
$= \operatorname{cosec} 35^{\circ} \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \sec 35^{\circ}$
Since $\operatorname{cosec} 35^{\circ} = \frac{1}{\sin 35^{\circ}}$ and $\sec 35^{\circ} = \frac{1}{\cos 35^{\circ}}$:
$= \left(\frac{1}{\sin 35^{\circ}}\right) \cdot \sin 35^{\circ} + \cos 35^{\circ} \cdot \left(\frac{1}{\cos 35^{\circ}}\right)$
$= 1 + 1 = 2$.

(D) Given expression: $(1-\cos \theta)(1+\cos \theta)$.
Using the algebraic identity $(a-b)(a+b) = a^2 - b^2$,we have:
$(1-\cos \theta)(1+\cos \theta) = 1^2 - \cos^2 \theta = 1 - \cos^2 \theta$.
According to the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we can write $1 - \cos^2 \theta = \sin^2 \theta$.
Therefore,the correct option is $D$.

(C) Given the equation: $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$.
We know the fundamental trigonometric identity: $\sin ^{2} \theta + \cos ^{2} \theta = 1$.
Substituting this into the equation,we get: $\tan ^{2} \theta = 1$.
Taking the square root on both sides: $\tan \theta = 1$ (considering the acute angle).
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.

(D) We know that $\tan \theta = \cot(90^{\circ} - \theta)$.
Using this identity:
$\tan 48^{\circ} = \tan(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$
$\tan 67^{\circ} = \tan(90^{\circ} - 23^{\circ}) = \cot 23^{\circ}$
Now,substitute these values into the expression:
$\tan 23^{\circ} \tan 42^{\circ} \tan 48^{\circ} \tan 67^{\circ} = \tan 23^{\circ} \tan 42^{\circ} \cot 42^{\circ} \cot 23^{\circ}$
Rearranging the terms:
$= (\tan 23^{\circ} \cot 23^{\circ}) \times (\tan 42^{\circ} \cot 42^{\circ})$
Since $\tan \theta \times \cot \theta = 1$:
$= 1 \times 1 = 1$

(D) Given: $\sec 4A = \operatorname{cosec}(A - 20^\circ)$
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: $90^\circ + 20^\circ = A + 4A$.
$110^\circ = 5A$.
$A = \frac{110^\circ}{5} = 22^\circ$.

(B) Given that $A+B+C=180^{\circ}$.
We can write $A+B = 180^{\circ}-C$.
Dividing both sides by $2$,we get $\frac{A+B}{2} = \frac{180^{\circ}-C}{2} = 90^{\circ}-\frac{C}{2}$.
Now,taking the tangent on both sides: $\tan \left(\frac{A+B}{2}\right) = \tan \left(90^{\circ}-\frac{C}{2}\right)$.
Using the trigonometric identity $\tan(90^{\circ}-\theta) = \cot \theta$,we get $\tan \left(\frac{A+B}{2}\right) = \cot \frac{C}{2}$.

(D) We know that $\sin(90^{\circ}-A) = \cos A$.
Therefore,$\sin(50^{\circ}+\theta) = \cos(90^{\circ}-(50^{\circ}+\theta))$.
$= \cos(90^{\circ}-50^{\circ}-\theta) = \cos(40^{\circ}-\theta)$.
Substituting this into the original expression:
$\cos(40^{\circ}-\theta) - \sin(50^{\circ}+\theta) = \cos(40^{\circ}-\theta) - \cos(40^{\circ}-\theta) = 0$.

(C) We know that $\cos(90^{\circ}-\theta) = \sin \theta$ and $\sin(90^{\circ}-\theta) = \cos \theta$.
Therefore,$\cos^{2} 50^{\circ} = \cos^{2}(90^{\circ}-40^{\circ}) = \sin^{2} 40^{\circ}$.
Similarly,$\sin^{2} 50^{\circ} = \sin^{2}(90^{\circ}-40^{\circ}) = \cos^{2} 40^{\circ}$.
Substituting these values into the expression:
$\frac{\cos^{2} 40^{\circ} + \sin^{2} 40^{\circ}}{\sin^{2} 40^{\circ} + \cos^{2} 40^{\circ}}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$\frac{1}{1} = 1$.

(A) We use the trigonometric identities for complementary angles:
$\cos (90^{\circ}- A ) = \sin A$
$\sin (90^{\circ}- A ) = \cos A$
$\tan (90^{\circ}- A ) = \cot A$
Substituting these into the expression:
$\frac{\cos (90^{\circ}- A ) \sin (90^{\circ}- A )}{\tan (90^{\circ}- A )} = \frac{\sin A \cdot \cos A}{\cot A}$
Since $\cot A = \frac{\cos A}{\sin A}$,we have:
$= \frac{\sin A \cdot \cos A}{\frac{\cos A}{\sin A}}$
$= \sin A \cdot \cos A \cdot \frac{\sin A}{\cos A}$
$= \sin A \cdot \sin A = \sin ^{2} A$

(C) Given that $\sin 3 \theta = \cos (\theta - 26^{\circ})$.
We know that $\sin A = \cos (90^{\circ} - A)$.
Therefore,$\cos (90^{\circ} - 3 \theta) = \cos (\theta - 26^{\circ})$.
Equating the angles,we get $90^{\circ} - 3 \theta = \theta - 26^{\circ}$.
Rearranging the terms,$90^{\circ} + 26^{\circ} = \theta + 3 \theta$.
$116^{\circ} = 4 \theta$.
$\theta = \frac{116^{\circ}}{4} = 29^{\circ}$.
Thus,the value of $\theta$ is $29^{\circ}$.

(C) Given that $\tan A = \cot B$.
We know the trigonometric identity $\cot B = \tan(90^\circ - B)$.
Substituting this into the equation,we get $\tan A = \tan(90^\circ - B)$.
By comparing the angles,we have $A = 90^\circ - B$.
Rearranging the terms,we get $A + B = 90^\circ$.

(D) Given expression: $2 \sin ^{2} \theta+4 \sec ^{2} \theta+5 \cot ^{2} \theta+2 \cos ^{2} \theta-4 \tan ^{2} \theta-5 \operatorname{cosec}^{2} \theta$
Rearranging the terms based on trigonometric identities:
$= 2(\sin ^{2} \theta + \cos ^{2} \theta) + 4(\sec ^{2} \theta - \tan ^{2} \theta) - 5(\operatorname{cosec}^{2} \theta - \cot ^{2} \theta)$
Using the fundamental trigonometric identities:
$1) \sin ^{2} \theta + \cos ^{2} \theta = 1$
$2) \sec ^{2} \theta - \tan ^{2} \theta = 1$
$3) \operatorname{cosec}^{2} \theta - \cot ^{2} \theta = 1$
Substituting these values into the expression:
$= 2(1) + 4(1) - 5(1)$
$= 2 + 4 - 5$
$= 1$

(A) We know the trigonometric identity: $\sin^2 \theta + \cos^2 \theta = 1$.
Therefore,$\sin^2 \theta = 1 - \cos^2 \theta$.
Substituting the given value of $\cos \theta = \frac{b}{\sqrt{a^2 + b^2}}$:
$\sin^2 \theta = 1 - \left( \frac{b}{\sqrt{a^2 + b^2}} \right)^2 = 1 - \frac{b^2}{a^2 + b^2}$.
Taking the common denominator:
$\sin^2 \theta = \frac{(a^2 + b^2) - b^2}{a^2 + b^2} = \frac{a^2}{a^2 + b^2}$.
Since $0 < \theta < 90^\circ$,$\sin \theta$ must be positive.
Thus,$\sin \theta = \sqrt{\frac{a^2}{a^2 + b^2}} = \frac{a}{\sqrt{a^2 + b^2}}$.

(C) Given that $\sec \theta = \frac{5}{3}$.
We use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$.
Substituting the value of $\sec \theta$:
$\tan^2 \theta = \left(\frac{5}{3}\right)^2 - 1$
$\tan^2 \theta = \frac{25}{9} - 1$
$\tan^2 \theta = \frac{25 - 9}{9} = \frac{16}{9}$
Taking the square root on both sides:
$\tan \theta = \sqrt{\frac{16}{9}} = \frac{4}{3}$.

(A) We know that $\frac{1}{\cos ^{2} \theta} = \sec ^{2} \theta$ and $\frac{1}{\cot ^{2} \theta} = \tan ^{2} \theta$.
Substituting these into the expression,we get $\sec ^{2} \theta - \tan ^{2} \theta$.
Using the trigonometric identity $1 + \tan ^{2} \theta = \sec ^{2} \theta$,we have $\sec ^{2} \theta - \tan ^{2} \theta = 1$.

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

(A) Given equations are:
$a \sin \theta = 3$ --- $(1)$
$a \cos \theta = 4$ --- $(2)$
Squaring both equations:
$(a \sin \theta)^2 = 3^2 \implies a^2 \sin^2 \theta = 9$ --- $(3)$
$(a \cos \theta)^2 = 4^2 \implies a^2 \cos^2 \theta = 16$ --- $(4)$
Adding equations $(3)$ and $(4)$:
$a^2 \sin^2 \theta + a^2 \cos^2 \theta = 9 + 16$
$a^2 (\sin^2 \theta + \cos^2 \theta) = 25$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$a^2 (1) = 25$
$a^2 = 25$
Since $a > 0$,we take the positive square root:
$a = 5$.

(D) We know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
Substituting these into the expression: $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$.
Taking the common denominator: $\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$.
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we get: $\frac{1}{\sin \theta \cos \theta}$.
Since $\frac{1}{\sin \theta} = \operatorname{cosec} \theta$ and $\frac{1}{\cos \theta} = \sec \theta$,the expression simplifies to $\operatorname{cosec} \theta \cdot \sec \theta$.

(B) We are given the expression: $(\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2$.
Using the algebraic identities $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$,we expand the terms:
$= (\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta) + (\sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta)$.
Combining like terms,the $2 \sin \theta \cos \theta$ and $-2 \sin \theta \cos \theta$ cancel out:
$= \sin^2 \theta + \cos^2 \theta + \sin^2 \theta + \cos^2 \theta$.
$= 2 \sin^2 \theta + 2 \cos^2 \theta$.
$= 2(\sin^2 \theta + \cos^2 \theta)$.
Since the fundamental trigonometric identity is $\sin^2 \theta + \cos^2 \theta = 1$,we substitute this value:
$= 2(1) = 2$.

(A) We know the fundamental trigonometric identity: $\sin ^{2} A + \cos ^{2} A = 1$.
Comparing the given equation $\sin ^{2} 35^{\circ} + \cos ^{2} \theta = 1$ with the identity $\sin ^{2} A + \cos ^{2} A = 1$,we can see that the angles must be equal for the sum to be $1$.
Therefore,$\theta = 35^{\circ}$.

(B) Given the equation: $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$
We know the fundamental trigonometric identity: $\sin ^{2} \theta + \cos ^{2} \theta = 1$
Substituting this identity into the equation,we get: $\tan ^{2} \theta = 1$
Taking the square root on both sides,and considering $0 < \theta < 90^{\circ}$ (where $\tan \theta$ is positive): $\tan \theta = 1$
Since $\tan 45^{\circ} = 1$,we conclude that $\theta = 45^{\circ}$.

(B) Given the equation $\sin ^{2}(3 x+30^{\circ})+\cos ^{2}(2 x+45^{\circ})=1$.
We know the fundamental trigonometric identity $\sin ^{2} \theta+\cos ^{2} \theta=1$.
Comparing the given equation with the identity,we have $3 x+30^{\circ} = 2 x+45^{\circ}$.
Subtracting $2x$ from both sides,we get $x+30^{\circ} = 45^{\circ}$.
Subtracting $30^{\circ}$ from both sides,we get $x = 15^{\circ}$.

(D) We know the trigonometric identities for complementary angles: $\tan(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.
First,consider the term $\tan(65^\circ - \theta)$:
$\tan(65^\circ - \theta) = \cot(90^\circ - (65^\circ - \theta)) = \cot(25^\circ + \theta)$.
Next,consider the term $\sec(55^\circ - \theta)$:
$\sec(55^\circ - \theta) = \operatorname{cosec}(90^\circ - (55^\circ - \theta)) = \operatorname{cosec}(35^\circ + \theta)$.
Now,substitute these into the given expression:
$\tan(65^\circ - \theta) - \cot(25^\circ + \theta) - \sec(55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta)$
$= \cot(25^\circ + \theta) - \cot(25^\circ + \theta) - \operatorname{cosec}(35^\circ + \theta) + \operatorname{cosec}(35^\circ + \theta)$
$= 0$.

(C) We know that $\frac{1}{\cos \theta} = \sec \theta$.
Therefore,$\frac{1}{\cos ^{2} \theta} = \sec ^{2} \theta$.
Using the trigonometric identity $1 + \tan ^{2} \theta = \sec ^{2} \theta$,we can write $\sec ^{2} \theta - 1 = \tan ^{2} \theta$.
Thus,$\frac{1}{\cos ^{2} \theta} - 1 = \tan ^{2} \theta$.

(B) We know that $\frac{1}{\sin \theta} = \operatorname{cosec} \theta$.
Therefore,$\frac{1}{\sin ^{2} \theta} = \operatorname{cosec}^{2} \theta$.
Substituting this into the expression: $\operatorname{cosec}^{2} \theta - 1$.
Using the trigonometric identity $1 + \cot^{2} \theta = \operatorname{cosec}^{2} \theta$,we get $\operatorname{cosec}^{2} \theta - 1 = \cot^{2} \theta$.

(A) We know the fundamental trigonometric identity: $\sec ^{2} \theta - \tan ^{2} \theta = 1$.
To find the value of $\tan ^{2} \theta - \sec ^{2} \theta$,we can multiply the identity by $-1$.
$-1 \times (\sec ^{2} \theta - \tan ^{2} \theta) = -1 \times 1$.
Therefore,$\tan ^{2} \theta - \sec ^{2} \theta = -1$.

(B) We know the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.
Applying this to the expression $\sec ^{4} \theta-\tan ^{4} \theta$,we get:
$\sec ^{4} \theta-\tan ^{4} \theta = (\sec ^{2} \theta - \tan ^{2} \theta)(\sec ^{2} \theta + \tan ^{2} \theta)$.
From the fundamental trigonometric identity,we know that $\sec ^{2} \theta - \tan ^{2} \theta = 1$.
Substituting the given values:
$\sec ^{4} \theta - \tan ^{4} \theta = (1) \times \left(\frac{13}{12}\right) = \frac{13}{12}$.

(D) We are given the expression $\frac{\sec \theta-1}{\sec \theta+1}$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we can multiply the numerator and the denominator by $\cos \theta$:
$\frac{\sec \theta-1}{\sec \theta+1} = \frac{(\sec \theta-1) \cdot \cos \theta}{(\sec \theta+1) \cdot \cos \theta}$
$= \frac{\sec \theta \cdot \cos \theta - \cos \theta}{\sec \theta \cdot \cos \theta + \cos \theta}$
$= \frac{1 - \cos \theta}{1 + \cos \theta}$ (since $\sec \theta \cdot \cos \theta = 1$)
Thus,the correct option is $D$.

(D) We know that $\cot \theta = \frac{1}{\tan \theta}$.
Substituting this into the expression,we get:
$\cot \theta \cdot \tan \theta = \left( \frac{1}{\tan \theta} \right) \cdot \tan \theta = 1$.

(D) Given equation: $\tan 7 \theta \cdot \tan 3 \theta = 1$
We know that $\tan 3 \theta = \frac{1}{\cot 3 \theta}$,so $\tan 7 \theta = \frac{1}{\tan 3 \theta} = \cot 3 \theta$.
Using the trigonometric identity $\cot \alpha = \tan(90^{\circ} - \alpha)$,we can write:
$\tan 7 \theta = \tan(90^{\circ} - 3 \theta)$
Equating the angles:
$7 \theta = 90^{\circ} - 3 \theta$
Adding $3 \theta$ to both sides:
$10 \theta = 90^{\circ}$
Dividing by $10$:
$\theta = 9^{\circ}$

(C) Given expression: $\tan 5^{\circ} \cdot \tan 25^{\circ} \cdot \tan 45^{\circ} \cdot \tan 65^{\circ} \cdot \tan 85^{\circ}$
We know that $\tan 45^{\circ} = 1$ and $\tan(90^{\circ} - \theta) = \cot \theta$.
Substituting these values:
$= \tan 5^{\circ} \cdot \tan 25^{\circ} \cdot 1 \cdot \tan(90^{\circ} - 25^{\circ}) \cdot \tan(90^{\circ} - 5^{\circ})$
$= \tan 5^{\circ} \cdot \tan 25^{\circ} \cdot 1 \cdot \cot 25^{\circ} \cdot \cot 5^{\circ}$
Since $\tan \theta \cdot \cot \theta = 1$,we group the terms:
$= (\tan 5^{\circ} \cdot \cot 5^{\circ}) \cdot (\tan 25^{\circ} \cdot \cot 25^{\circ}) \cdot 1$
$= 1 \cdot 1 \cdot 1 = 1$

(A) We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Therefore,$\sin 75^{\circ} = \sin(90^{\circ} - 15^{\circ}) = \cos 15^{\circ}$.
Substituting this into the expression:
$\sin^{2} 15^{\circ} + \sin^{2} 75^{\circ} = \sin^{2} 15^{\circ} + (\cos 15^{\circ})^{2}$
$= \sin^{2} 15^{\circ} + \cos^{2} 15^{\circ}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$= 1$.

(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we can rewrite the terms:
$\sin ^{2} 87^{\circ} = \sin ^{2}(90^{\circ} - 3^{\circ}) = \cos ^{2} 3^{\circ}$
$\sin ^{2} 89^{\circ} = \sin ^{2}(90^{\circ} - 1^{\circ}) = \cos ^{2} 1^{\circ}$
Substituting these into the original expression:
$= \sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ} + \cos ^{2} 1^{\circ}$
Grouping the terms with the same angle:
$= (\sin ^{2} 1^{\circ} + \cos ^{2} 1^{\circ}) + (\sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ})$
Using the identity $\sin ^{2} \theta + \cos ^{2} \theta = 1$:
$= 1 + 1 = 2$.

(C) We know that the reciprocal of $\tan \theta$ is $\cot \theta$,so $\frac{1}{\tan \theta} = \cot \theta$.
Therefore,$\frac{1}{\tan^{2} \theta} = \cot^{2} \theta$.
Substituting this into the expression,we get $\cot^{2} \theta + 1$.
Using the trigonometric identity $1 + \cot^{2} \theta = \operatorname{cosec}^{2} \theta$,the final result is $\operatorname{cosec}^{2} \theta$.