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Question: If \(\cosh\alpha = \sec x,\) then \(\tan^{2}\frac{x}{2} =\)...

If cosh⁡α=sec⁡x,\cosh\alpha = \sec x, then tan⁡2x2=\tan^{2}\frac{x}{2} =

A

cos⁡2α2\cos^{2}\frac{\alpha}{2}

B

sin⁡2α2\sin^{2}\frac{\alpha}{2}

C

cot⁡2α2\cot^{2}\frac{\alpha}{2}

D

tan⁡h2α2\tan h^{2}\frac{\alpha}{2}

Answer

tan⁡h2α2\tan h^{2}\frac{\alpha}{2}

Explanation

Solution

cos⁡hα=sec⁡x=1cos⁡x=11−tan⁡2x21+tan⁡2x2{\cos h}\alpha = \sec x = \frac{1}{\cos x} = \frac{1}{\frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}

cos⁡hα1=1+tan⁡2x21−tan⁡2x2\frac{\cos h\alpha}{1} = \frac{1 + \tan^{2}\frac{x}{2}}{1 - \tan^{2}\frac{x}{2}} ⇒\Rightarrow cos⁡hα−1cos⁡hα+1=2tan⁡2x22=tan⁡2x2\frac{{\cos h}\alpha - 1}{{\cos h}\alpha + 1} = \frac{2\tan^{2}\frac{x}{2}}{2} = \tan^{2}\frac{x}{2}

⇒\Rightarrow 2sin⁡h2α22cos⁡h2α2=tan⁡2x2\frac{2\sin h^{2}\frac{\alpha}{2}}{2\cos h^{2}\frac{\alpha}{2}} = \tan^{2}\frac{x}{2} ⇒\Rightarrow tan⁡h2α2=tan⁡2x2{\tan h}^{2}\frac{\alpha}{2} = \tan^{2}\frac{x}{2}