Question : If $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=\frac{3}{2}$, then the value of $\sin ^4 \theta-\cos ^4 \theta$ is:
Option 1: $\frac{5}{12}$
Option 2: $\frac{12}{13}$
Option 3: $\frac{11}{12}$
Option 4: $\frac{5}{13}$
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Correct Answer: $\frac{12}{13}$
Solution :
Given, $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=\frac{3}{2}$
or, $2(\sin \theta+\cos \theta)=3(\sin \theta-\cos \theta)$
or, $5\cos \theta = \sin \theta$
We know that $\sin^2 A+\cos^2 A=1$
Using $5\cos \theta = \sin \theta$, we get
$25\cos^2 \theta +\cos^2 \theta=1$
or, $\cos^2 \theta = \frac{1}{26}$
Since $(a^2-b^2)=(a-b)(a+b)$
$\sin ^4 \theta-\cos ^4 \theta = (\sin ^2 \theta-\cos ^2 \theta$)($\sin ^2 \theta+\cos ^2 \theta$)
or, $\sin ^4 \theta-\cos ^4 \theta= (\sin ^2 \theta-\cos ^2 \theta$)
or, $\sin ^4 \theta-\cos ^4 \theta= 25\cos ^2 \theta-\cos ^2 \theta$
or, $\sin ^4 \theta-\cos ^4 \theta= 24\cos ^2 \theta$
or, $\sin ^4 \theta-\cos ^4 \theta= 24\times\frac{1}{26}$
or, $\sin ^4 \theta-\cos ^4 \theta= 12\times\frac{1}{13}$
Hence, the correct answer is $\frac{12}{13}$.
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