If $\cos ^{-1} \sqrt{\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{q}}=\frac{3 \pi}{4}$, then $\mathrm{q}$

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Aicharya · Accepted Answer

$\begin{aligned} & \cos ^{-1} \sqrt{\mathrm{p}}-\cos ^{-1} \sqrt{1-\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{q}}=\frac{3 \pi}{4} \ & 	ext { Let } \mathrm{t}=\cos ^{-1} \sqrt{\mathrm{p}} \ & \Rightarrow \mathrm{p}=\cos ^2 \mathrm{t} \ & \Rightarrow \mathrm{p}=1-\sin ^2 \mathrm{t} \ & \Rightarrow \sin \mathrm{t}=\sqrt{1-\mathrm{p}} \ & \Rightarrow \mathrm{t}=\sin ^{-1} \sqrt{1-\mathrm{p}}\end{aligned}$ $\Rightarrow \cos ^{-1} \sqrt{\mathrm{p}}=\sin ^{-1} \sqrt{1-\mathrm{p}}$ $	herefore \quad$ Given equation becomes $\sin ^{-1} \sqrt{1-p}-\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ $	herefore \quad \frac{\pi}{2}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4} \quad \ldots\left[\because \cos ^{-1} a+\sin ^{-1} a=\frac{\pi}{2}ight]$ $\begin{array}{ll}	herefore & \cos ^{-1} \sqrt{1-q}=\frac{-\pi}{4} \ 	herefore & \sqrt{1-q}=\cos \left(-\frac{\pi}{4}ight) \ 	herefore & q=1-\frac{1}{2} \ 	herefore & q=\frac{1}{2}\end{array}$

If $\cos ^{-1} \sqrt{\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{p}}+\cos ^{-1} \sqrt{1-\mathrm{q}}=\frac{3…

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