Question

If $tan\left(co{s}^{-1}x\right)=sin\left(co{t}^{-1}\frac{1}{2}\right),$ then x =

• A.
  
• B.
• C.
• D. None of these

Answer 1

The correct option is B

Given that $tan\left\{co{s}^{-1}\right(x\left)\right\}=sin\left(co{t}^{-1}\frac{1}{2}\right)$
$Letco{t}^{-1}\frac{1}{2}=\varphi \Rightarrow \frac{1}{2}=cot\varphi$
$\Rightarrow sin\varphi =\frac{1}{\sqrt{1+co{t}^2\varphi }}=\frac{2}{\sqrt{5}}$
$Letco{s}^{-1}x=\theta \Rightarrow sec\theta \frac{1}{x}\Rightarrow tan\theta =\sqrt{se{c}^2\theta -1}$
$\Rightarrow tan\theta =\sqrt{\frac{1}{x^2}-1}\Rightarrow tan\theta =\frac{\sqrt{1-x^2}}{x}$
$So,tan\left\{co{s}^{-1}\right(x\left)\right\}=sin\left(co{t}^{-1}\frac{1}{2}\right)$
$\Rightarrow tan\left(ta{n}^{-1}\frac{\sqrt{1-x^2}}{x}\right)=sin\left(si{n}^{-1}\frac{2}{\sqrt{5}}\right)\Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}}\Rightarrow \sqrt{(1-x^2)5}=2x$
Squaring both sides, we get $x=\pm \frac{\sqrt{5}}{3}.$