If sin theta cosh alpha = tan x and cos theta | vedclass

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(C) Given equations are: $\sin 	heta \cosh \alpha = 	an x$ and $\cos 	heta \sinh \alpha = \sec x$.Squaring and subtracting the equations: $\sec^2 x

Vedclass · Accepted Answer

$3$

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(C) Given equations are: $\sin 	heta \cosh \alpha = 	an x$ and $\cos 	heta \sinh \alpha = \sec x$.Squaring and subtracting the equations: $\sec^2 x - 	an^2 x = (\cos 	heta \sinh \alpha)^2 - (\sin 	heta \cosh \alpha)^2$.Since $\sec^2 x - 	an^2 x = 1$,we have: $\cos^2 	heta \sinh^2 \alpha - \sin^2 	heta \cosh^2 \alpha = 1$.Using $\sin^2 	heta = 1 - \cos^2 	heta$ and $\sinh^2 \alpha = \cosh^2 \alpha - 1$:$\cos^2 	heta (\cosh^2 \alpha - 1) - (1 - \cos^2 	heta) \cosh^2 \alpha = 1$.$\cos^2 	heta \cosh^2 \alpha - \cos^2 	heta - \cosh^2 \alpha + \cos^2 	heta \cosh^2 \alpha = 1$.$2 \cos^2 	heta \cosh^2 \alpha - \cos^2 	heta - \cosh^2 \alpha = 1$.Using the identity $\cosh 2 \alpha = 2 \cosh^2 \alpha - 1 \implies \cosh^2 \alpha = \frac{\cosh 2 \alpha + 1}{2}$ and $\cos 2 	heta = 2 \cos^2 	heta - 1 \implies \cos^2 	heta = \frac{\cos 2 	heta + 1}{2}$.Substituting these: $2 \left( \frac{\cos 2 	heta + 1}{2} ight) \left( \frac{\cosh 2 \alpha + 1}{2} ight) - \left( \frac{\cos 2 	heta + 1}{2} ight) - \left( \frac{\cosh 2 \alpha + 1}{2} ight) = 1$.Multiplying by $2$: $(\cos 2 	heta + 1)(\cosh 2 \alpha + 1) - (\cos 2 	heta + 1) - (\cosh 2 \alpha + 1) = 2$.$\cos 2 	heta \cosh 2 \alpha + \cos 2 	heta + \cosh 2 \alpha + 1 - \cos 2 	heta - 1 - \cosh 2 \alpha - 1 = 2$.$\cos 2 	heta \cosh 2 \alpha - 1 = 2$.Therefore,$\cos 2 	heta \cosh 2 \alpha = 3$.

If $\sin \theta \cosh \alpha = \tan x$ and $\cos \theta \sinh \alpha = \sec x$,then find the value of $\cos 2 \theta \cosh 2 \alpha$.

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