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If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$, then find the value of $x$.
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$\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=\sin \frac{\pi}{2}$ or $\sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$
or $\left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} \frac{1}{5}\right)+\left(-\cos ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=\frac{\pi}{2}$
or $\frac{\pi}{2}-\cos ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$
or $\cos ^{-1} \frac{1}{5}-\cos ^{-1} x=0$ or $\cos ^{-1} x=\cos ^{-1} \frac{1}{5}$ $\Rightarrow \quad x=\frac{1}{5}$
or $\left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} \frac{1}{5}\right)+\left(-\cos ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=\frac{\pi}{2}$
or $\frac{\pi}{2}-\cos ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$
or $\cos ^{-1} \frac{1}{5}-\cos ^{-1} x=0$ or $\cos ^{-1} x=\cos ^{-1} \frac{1}{5}$ $\Rightarrow \quad x=\frac{1}{5}$
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