• # question_answer If ${{\sin }^{-1}}\frac{x}{5}+\cos e{{c}^{-1}}\frac{5}{4}=\frac{\pi }{2}$, then $x$ is equal to : A)  1                                             B)  4 C)  3                                             D)  5

We have, ${{\sin }^{-1}}\frac{x}{5}+\cos e{{c}^{-1}}\frac{5}{4}=\frac{\pi }{2}$ $\Rightarrow$               ${{\sin }^{-1}}\frac{x}{5}+{{\sin }^{-1}}\frac{4}{5}=\frac{\pi }{2}$ $\Rightarrow \,{{\sin }^{-1}}\left[ \frac{x}{5}\times \sqrt{1-{{(4/5)}^{2}}}+\frac{4}{5}\times \sqrt{1-{{(x/5)}^{2}}} \right]=\pi /2$or $\frac{x}{5}\times \frac{3}{5}+\frac{4}{5}\times \frac{\sqrt{25-{{x}^{2}}}}{5}=\sin \pi /2$ $\Rightarrow \,\,3x+4\sqrt{25-{{x}^{2}}}=25$ $\Rightarrow$               $4\sqrt{25-{{x}^{2}}}=25-3x$ Squaring both sides, we get                 $16\,(25-{{x}^{2}})=625+9{{x}^{2}}-150x$ $\Rightarrow \,\,\,4500-16{{x}^{2}}=625+9{{x}^{2}}-150x$ $\Rightarrow \,\,\,\,16{{x}^{2}}+9{{x}^{2}}-150x+625-400=0$ $\Rightarrow$               $25{{x}^{2}}-150x+225=0$ $\Rightarrow$                               ${{x}^{2}}-6x+9=0$ or            ${{x}^{2}}-3x-3x+9=0$                 $x\left( x-3 \right)-3\left( x-3 \right)=0$ $\Rightarrow$                               ${{\left( x-3 \right)}^{2}}=0$ $\therefore$                                  $x=3$