Question

if ${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6},thenxis:$

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $1$
• D. none of these

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

Solution :- given

$\Rightarrow si{n}^{-1}x-co{s}^{-1}x=\frac{\pi }{6}$ (We know $\Rightarrow si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}x)$

$\Rightarrow si{n}^{-1}x-\frac{\pi }{2}+si{n}^{-1}x=\frac{\pi }{6}$

$\Rightarrow 2si{n}^{-1}x=\frac{\pi }{6}+\frac{\pi }{2}$

$\Rightarrow 2si{n}^{-1}x=\frac{\pi +3\pi }{6}$

$\Rightarrow si{n}^{-1}x=\frac{4\pi }{6\times 2}$

$\Rightarrow si{n}^{-1}x=\frac{\pi }{3}$

$\Rightarrow x=sin\left(\frac{\pi }{3}\right)$

$\Rightarrow x=\frac{\sqrt{3}}{2}$ Ans