Question

Find the value of expression tan $\bigg(\frac{sin^{-1} x+ cos^{-1}x}{2} \bigg) , when \ x=\frac{\sqrt{3}}{2}$

Answer 1

$\underline{\underline{\text{Solution :}}}$

$\underline{\text{Method 1}}$

$\mathrm{Now,\:\frac{sin^{-1}x+cos^{-1}x}{2}}$

$\mathrm{=\frac{sin^{-1}\frac{\sqrt{3}}{2}+cos^{-1}\frac{\sqrt{3}}{2}}{2}}$

$\mathrm{=\frac{\frac{\pi}{3}+\frac{\pi}{6}}{2}}$

$\mathrm{=\frac{1}{2}*\frac{2\pi+\pi}{2}}$

$\mathrm{=\frac{1}{2}*\frac{3\pi}{6}}$

$\mathrm{=\frac{\pi}{4}}$

$\to \boxed{\mathrm{\frac{sin^{-1}x+cos^{-1}x}{2}=\frac{\pi}{4}}}$

$\underline{\text{Method 2}}$

$\mathrm{Now,\:\frac{sin^{-1}x+cos^{-1}x}{2}}$

$\mathrm{=\frac{1}{2}*\frac{\pi}{2}}$

$\mathrm{=\frac{\pi}{4}}$

$\to \boxed{\mathrm{\frac{sin^{-1}x+cos^{-1}x}{2}=\frac{\pi}{4}}}$

$\underline{\text{Rules :}}$

$\mathrm{1.\:sin^{-1}x+cos^{-1}x=\frac{\pi}{2}}$

$\mathrm{2.\:sin^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{3}}$

$\mathrm{3.\:cos^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{6}}$

$\underline{\text{Extra :}}$

$\text{We know that,}$

$\mathrm{sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}}$

$\to \mathrm{\frac{\pi}{3}=sin^{-1}\frac{\sqrt{3}}{2}}$

$\to \mathrm{sin^{-1}\frac{\sqrt{3}}{2}=\frac{\pi}{3}}$