Question : If $\sin A+\sin ^2 A=1$, then the value of the expression $\left(\cos ^2 A+\cos ^4 A\right)$ is
Option 1: $\frac{3}{2}$
Option 2: $1$
Option 3: $2$
Option 4: $\frac{1}{2}$
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Correct Answer: $1$
Solution : Given, $\sin A+\sin ^2 A=1$ $⇒\sin A=1-\sin ^2 A$ $⇒\sin A=\cos ^2 A$ So, $\cos ^2 A+\cos ^4 A= \cos ^2 A+\sin ^2 A=1$ [$\because \cos^2 A + \sin^2A=1$] Hence, the correct answer is 1.
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Question : What is the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}$?
Option 1: $\frac{1}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 2: $2\left(\sin ^2 \theta-\cos ^2 \theta\right)$
Option 3: $\frac{2}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 4: $\sin ^2 \theta-\cos ^2 \theta$
Question : If $\sin \theta \cos \theta=\frac{1}{\sqrt{3}}$ then the value of $\left(\sin ^4 \theta+\cos ^4 \theta\right)$ is:
Option 1: $1$
Option 2: $\frac{5}{3}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{1}{3}$
Question : If $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$,then the value of $\left(\sin ^6 \theta+\cos ^6 \theta\right)$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{4}{3}$
Option 4: $\frac{5}{3}$
Question : If $x=a\left ( \sin\theta+\cos\theta \right )$ and $y=b\left ( \sin\theta-\cos\theta \right )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:
Option 1: 4
Option 2: 3
Option 3: 1
Option 4: 2
Question : Simplify the given equation: $(1+\tan ^2 A)(1+\cot ^2 A)=?$
Option 1: $\frac{1}{\cos ^2 A\left(1+\sin ^2 A\right)}$
Option 2: $\frac{1}{\sin ^2 A\left(1-\sin ^2 A\right)}$
Option 3: $\frac{1}{\sin ^2 A+\operatorname{cosec}^2 A}$
Option 4: $\frac{1}{\sin ^2 A\left(1+\cos ^2 A\right)}$
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