Question

The value of x for which $sin\left(co{t}^{-1}\right(1+x\left)\right)=cos\left(ta{n}^{-1}x\right)$ is:

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

Answer 1

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

$sin\left(co{t}^{-1}\right(1+x\left)\right)=cos\left(ta{n}^{-1}x\right)\Rightarrow co{t}^{-1}(1+x)=[\frac{\pi }{2}\pm ta{n}^{-1}x]\Rightarrow \frac{\pi }{2}-ta{n}^{-1}(1+x)=\frac{\pi }{2}\pm ta{n}^{-1}x\Rightarrow 1+x=\pm x$
$x=-\frac{1}{2}$ Which, on verification, satisfies the equation.