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Question

If sinxsinhy=cosθ and cosxcoshy=sinθ then cosh2y+cos2x=

A
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Solution

The correct option is C 2
sinxsinhy=cosθ

Squaring both sides,
sin2xsinh2y=cos2θ ....................... (1)

And,
cosxcoshy=sinθ

Squaring both sides,
cos2xcosh2y=sin2θ ....................... (2)

Adding (1) and (2), we get,

sin2xsinh2y+cos2xcosh2y=cos2θ+sin2θ

sin2xsinh2y+(1sin2x)cosh2y=1

sin2xsinh2y+cosh2ysin2xcosh2y=1

sin2x(sinh2ycosh2y)+cosh2y=1

sin2x+cosh2y=1 (cosh2ysinh2y=1)

Add 1 both sides,
1sin2x+cosh2y=1+1

cos2x+cosh2y=2

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