Solveeit Logo
Login

Question

Question: If \(u = {logtan}\left( \frac{\pi}{4} + \frac{x}{2} \right),\) then \(\cos ⥂ hu\) is equal to...

If u=logtan(π4+x2),u = {logtan}\left( \frac{\pi}{4} + \frac{x}{2} \right), then cos⁡⥂hu\cos ⥂ hu is equal to

A

sec⁡x\sec x

B

cosecxc\text{osec}x

C

tan⁡x\tan x

D

sin⁡x\sin x

Answer

sec⁡x\sec x

Explanation

Solution

u=logtan(π4+x2)u = {logtan}\left( \frac{\pi}{4} + \frac{x}{2} \right) ⇒ ∴\therefore eu1=1+tan⁡x21−tan⁡x2\frac{e^{u}}{1} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}

cosh⁡u=eu+e−u2=e2u+12eu=(1+tan⁡x21−tan⁡x2)2+12.(1+tan⁡x21−tan⁡x2)\cosh u = \frac{e^{u} + e^{- u}}{2} = \frac{e^{2u} + 1}{2e^{u}} = \frac{\left( \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} \right)^{2} + 1}{2.\left( \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} \right)}=2[1+tan⁡2x2]2(1−tan⁡x2)(1+tan⁡x2)=1+tan⁡2π21−tan⁡2π2\frac{2\left\lbrack 1 + \tan^{2}\frac{x}{2} \right\rbrack}{2\left( 1 - \tan\frac{x}{2} \right)\left( 1 + \tan\frac{x}{2} \right)} = \frac{1 + \tan^{2}\frac{\pi}{2}}{1 - \tan^{2}\frac{\pi}{2}}= 11−tan⁡2π21+tan⁡2π2=1cos⁡x=sec⁡x\frac{1}{\frac{1 - \tan^{2}\frac{\pi}{2}}{1 + \tan^{2}\frac{\pi}{2}}} = \frac{1}{\cos x} = \sec x