Question

The value of $x$ which satisfies $\sin \left({\cot }^{-1}x\right)=\cos \left({\tan }^{-1}\right(1+x))$ is ?

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-\frac{1}{2}$

$co{t}^{-1}x=si{n}^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)$ ...[1]

$ta{n}^{-1}x=co{s}^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)$

$⟹ta{n}^{-1}(1+x)=co{s}^{-1}\left(\frac{1}{\sqrt{(1+x{)}^2+1}}\right)$ ...[2]

$sin\left(co{t}^{-1}x\right)=cos\left(ta{n}^{-1}\right(1+x\left)\right))$

$⟹sin\left(si{n}^{-1}\right(\frac{1}{\sqrt{x^2+1}}\left)\right)=cos\left(co{s}^{-1}\right(\frac{1}{\sqrt{(1+x{)}^2+1}}\left)\right)$ ...(from [1] and[2])

$⟹\frac{1}{\sqrt{x^2+1}}=\frac{1}{\sqrt{(1+x{)}^2+1}}$

$⟹2x+1=0$

$⟹x=\frac{-1}{2}$

Hence, the answer is option (A).