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Question

Question: If \(\cos ^ { - 1 } \sqrt { p } + \cos ^ { - 1 } \sqrt { 1 - p } + \cos ^ { - 1 } \sqrt { 1 - q } =...

If cos1p+cos11p+cos11q=3π4\cos ^ { - 1 } \sqrt { p } + \cos ^ { - 1 } \sqrt { 1 - p } + \cos ^ { - 1 } \sqrt { 1 - q } = \frac { 3 \pi } { 4 } then the value of q is.

A

1

B

12\frac { 1 } { \sqrt { 2 } }

C

13\frac { 1 } { 3 }

D

12\frac { 1 } { 2 }

Answer

12\frac { 1 } { 2 }

Explanation

Solution

Let α=cos1p\alpha = \cos ^ { - 1 } \sqrt { p } β=cos11p\beta = \cos ^ { - 1 } \sqrt { 1 - p }

And γ=cos11q\gamma = \cos ^ { - 1 } \sqrt { 1 - q } or cosα=p;cosβ=1p\cos \alpha = \sqrt { p } ; \cos \beta = \sqrt { 1 - p }

and

Thereforesinα=1p\sin \alpha = \sqrt { 1 - p } sinβ=p\sin \beta = \sqrt { p } and sinγ=q\sin \gamma = \sqrt { q }.

The given equation may be written as

α+β+γ=3π4\alpha + \beta + \gamma = \frac { 3 \pi } { 4 } or α+β=3π4γ\alpha + \beta = \frac { 3 \pi } { 4 } - \gamma or

cos(α+β)=cos(3π4γ)\cos ( \alpha + \beta ) = \cos \left( \frac { 3 \pi } { 4 } - \gamma \right)

cosαcosβsinαsinβ=\cos \alpha \cos \beta - \sin \alpha \sin \beta = cos{π(π4+γ)}=cos(π4+γ)\cos \left\{ \pi - \left( \frac { \pi } { 4 } + \gamma \right) \right\} = - \cos \left( \frac { \pi } { 4 } + \gamma \right)

=(121q12q)= - \left( \frac { 1 } { \sqrt { 2 } } \sqrt { 1 - q } - \frac { 1 } { \sqrt { 2 } } \cdot \sqrt { q } \right)

1q=q1 - q = q