Question : If $\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}$, then the value of $\frac{1+\cot ^2 x}{1-\cot ^2 x}$ is:
Option 1: 2.25
Option 2: 1.45
Option 3: 3.75
Option 4: 5.25
Correct Answer: 1.45
Solution :
$\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{2}{5}$
Dividing both the numerator and denominator by $\cos x$
⇒ $\frac{1-\cot x}{1+\cot x}=\frac{2}{5}$
⇒ $5-5\cot x=2+2\cot x$
⇒ $\cot x=\frac{3}{7}$
So, $\frac{1+\cot ^2 x}{1-\cot ^2 x}$=$\frac{1+(\frac{3}7) ^2}{1-(\frac{3}7)^2 }=\frac{58}{40} = 1.45$
Hence, the correct answer is 1.45.
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