Question

If ${\cos }^{-1}a+{\cos }^{-1}b+{\cos }^{-1}c=\pi$, then value of $(a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2})$ will be

• A. $2\sqrt{1-a^2}\sqrt{1-b^2}\sqrt{1-c^2}$
• B. $\sqrt{1-a^2}\sqrt{1-b^2}\sqrt{1-c^2}$
• C. $abc$
• D. $2abc$

Answer 1

The correct option is A $\left(2\sqrt{1-a^2}\sqrt{1-b^2}\sqrt{1-c^2}\right)$

Given, ${\cos }^{-1}a+{\cos }^{-1}b+{\cos }^{-1}c=\pi$

let ${\cos }^{-1}a=A\Rightarrow a=\cos A$

${\cos }^{-1}b=B\Rightarrow b=\cos B$

${\cos }^{-1}c=C\Rightarrow c=\cos C$
Substitute above values in given expression,
we get $A+B+C=\pi$ ....(i)

We, know In a triangle

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$

$\Rightarrow 2\sin A\cos A+2\sin B\cos B+2\sin C\cos C=4\sin A\sin B\sin C$

$\Rightarrow \sin A\cos A+\sin B\cos B+\sin C\cos C=2\sin A\sin B\sin C$
$\Rightarrow \cos A\sqrt{1-{\cos }^{2}A}+\cos B\sqrt{1-{\cos }^{2}B}+\cos C\sqrt{1-{\cos }^{2}C}=2\sqrt{1-{\cos }^{2}A}\sqrt{1-{\cos }^{2}B}\sqrt{1-{\cos }^{2}C}$ As ${\sin }^2\theta +{\cos }^2\theta =1$

$\therefore a\sqrt{1-a^2}+b\sqrt{1-b^2}+c\sqrt{1-c^2}=2\sqrt{1-a^2}\sqrt{1-b^2}\sqrt{1-c^2}$