sec left(tan ^{-1} frac{y}{2}right) = | vedclass

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Question

(C) Let $	heta = 	an ^{-1} \left(\frac{y}{2}ight)$.Then,$	an 	heta = \frac{y}{2}$.We know the identity $\sec ^2 	heta = 1 + 	an ^2 	heta$.Sub

Vedclass · Accepted Answer

$\frac{\sqrt{4+y^2}}{2}$

Vedclass · Answer

(C) Let $	heta = 	an ^{-1} \left(\frac{y}{2}ight)$.Then,$	an 	heta = \frac{y}{2}$.We know the identity $\sec ^2 	heta = 1 + 	an ^2 	heta$.Substituting the value of $	an 	heta$:$\sec ^2 	heta = 1 + \left(\frac{y}{2}ight)^2 = 1 + \frac{y^2}{4} = \frac{4+y^2}{4}$.Taking the square root on both sides:$\sec 	heta = \sqrt{\frac{4+y^2}{4}} = \frac{\sqrt{4+y^2}}{2}$.Thus,$\sec \left(	an ^{-1} \frac{y}{2}ight) = \frac{\sqrt{4+y^2}}{2}$.

$\sec \left(\tan ^{-1} \frac{y}{2}\right) = $

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