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Question: If \(\sin ^ { - 1 } x = \theta + \beta\)and \(1 + x y =\)...

If sin⁡−1x=θ+β\sin ^ { - 1 } x = \theta + \betaand 1+xy=1 + x y =

A

sin⁡2θ+sin⁡2β\sin ^ { 2 } \theta + \sin ^ { 2 } \beta

B

sin⁡2θ+cos⁡2β\sin ^ { 2 } \theta + \cos ^ { 2 } \beta

C

cos⁡2θ+cos⁡2β\cos ^ { 2 } \theta + \cos ^ { 2 } \beta

D

cos⁡2θ+sin⁡2β\cos ^ { 2 } \theta + \sin ^ { 2 } \beta

Answer

sin⁡2θ+cos⁡2β\sin ^ { 2 } \theta + \cos ^ { 2 } \beta

Explanation

Solution

Obviously x=sin⁡(θ+β)x = \sin ( \theta + \beta ) and

y=sin⁡(θ−β)y = \sin ( \theta - \beta )

∴1+xy=1+sin⁡(θ+β)sin⁡(θ−β)\therefore 1 + x y = 1 + \sin ( \theta + \beta ) \sin ( \theta - \beta )

=1+sin⁡2θ−sin⁡2β=sin⁡2θ+cos⁡2β= 1 + \sin ^ { 2 } \theta - \sin ^ { 2 } \beta = \sin ^ { 2 } \theta + \cos ^ { 2 } \beta.