Question : If $\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{4}{5}$, then the value of $\frac{\operatorname{cosec}^2 \theta}{2-\operatorname{cosec}^2 \theta}$ is:
Option 1: $\frac{16}{25}$
Option 2: $\frac{40}{41}$
Option 3: $\frac{41}{40}$
Option 4: $\frac{31}{30}$
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Correct Answer: $\frac{41}{40}$
Solution :
Given: $\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{4}{5}$
⇒ $5\sin \theta - 5\cos \theta = 4\sin \theta + 4\cos \theta$
$\sin \theta = 9\cos \theta$
Dividing both sides by $\cos \theta$, we get:
$\tan \theta = 9$
Now, we know that:
$\tan\theta=\frac{p}{b}=\frac{9}{1}$
So, $h=\sqrt{p^2+b^2}=\sqrt{9^2+1^2}=\sqrt{82}$
Now, $\frac{\operatorname{cosec}^2 \theta}{2-\operatorname{cosec}^2 \theta}$
⇒ $\frac{(\frac{h}{p})^{2}}{2-(\frac{h}{p})^{2}}$
Putting the values of $p$, $b$ and $h$, we get:
= $\frac{\frac{82}{81}}{2-\frac{82}{81}}$
= $\frac{\frac{82}{81}}{\frac{162-82}{81}}$
= $\frac{82}{80}$
= $\frac{41}{40}$
Hence, the correct answer is $\frac{41}{40}$.
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