Question

If $0<x<1$, then find the value of $\frac{x}{1-x}(\left(\sin \right({\cot }^{-1}x\left)\sec \right({\cot }^{-1}x){)}^2-1{)}^{\frac{1}{2}}$

• A. $\sqrt{\frac{1-x}{1+x}}$
• B. $\sqrt{\frac{1+x}{1-x}}$
• C. $\sqrt{\frac{1-x^2}{1+x^2}}$
• D. $\sqrt{1+x^2}$

Answer 1

The correct option is B $\sqrt{\frac{1+x}{1-x}}$

$\frac{x}{1-x}(\sin \left({\cot }^{-1}x\right)\sec \left({\cot }^{-1}x\right){)}^2-1{)}^{\frac{1}{2}}=\frac{x}{1-x}{({(\frac{1}{\sqrt{1+x^2}}\times \frac{\sqrt{1+x^2}}{x})}^2-1)}^{\frac{1}{2}}=\frac{x}{1-x}\times \sqrt{\frac{1-x^2}{x^2}}=\sqrt{\frac{1+x}{1-x}}$