Question : If $\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$, then the value of $\sin^{4}\theta$ is:
Option 1: $\frac{2}{5}$
Option 2: $\frac{1}{5}$
Option 3: $\frac{16}{25}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{16}{25}$
Solution :
Given that $\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=3$
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=\frac{3}{1}$
Applying Componendo and Dividendo,
$⇒\frac{2\sin\theta}{2\cos\theta} = \frac{4}{2}$
$⇒\frac{\sin\theta}{\cos\theta} = 2$
$⇒\tan\theta = \frac{2}{1}$
We know that, $\tan\theta =\frac{\text{Perpendicular}}{\text{Base}}$
Let be perpendicular = $2k$, and base = $k$.
Using Pythagoras theorem, hypotenuse = $\sqrt{(2k)^2+k^2}=\sqrt5k$
$\sin\theta =\frac{2}{\sqrt5}$
⇒ $\sin^{4}\theta=(\frac{2}{\sqrt5})^4=\frac{16}{25}$
Hence, the correct answer is $\frac{16}{25}$.
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