Question Number 211492 by swargiya last updated on 11/Sep/24
Commented by MathematicalUser2357 last updated on 12/Sep/24
$$ \:\mathrm{sin}^{−\mathrm{1}} {x}−\mathrm{cos}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{6}}\: ,\: \:{x}\: \: \: \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{math}\:\mathrm{question}\:\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{hindi}\:\mathrm{question} \\ $$
Answered by BHOOPENDRA last updated on 11/Sep/24
$${sin}^{−\mathrm{1}} {x}\:−\left(\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} {x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{2}{sin}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{3}} \\ $$$${x}={sin}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$