sec 2 theta - tan 2 theta = | vedclass

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(A) $\sec 2 	heta - 	an 2 	heta = \frac{1}{\cos 2 	heta} - \frac{\sin 2 	heta}{\cos 2 	heta} = \frac{1 - \sin 2 	heta}{\cos 2 	heta}$Using the

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$	an \left(\frac{\pi}{4} - 	hetaight)$

Vedclass · Answer

(A) $\sec 2 	heta - 	an 2 	heta = \frac{1}{\cos 2 	heta} - \frac{\sin 2 	heta}{\cos 2 	heta} = \frac{1 - \sin 2 	heta}{\cos 2 	heta}$Using the identities $1 - \sin 2 	heta = (\cos 	heta - \sin 	heta)^2$ and $\cos 2 	heta = \cos^2 	heta - \sin^2 	heta = (\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta)$:$\frac{(\cos 	heta - \sin 	heta)^2}{(\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta)} = \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta}$Dividing numerator and denominator by $\cos 	heta$:$\frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{	an \frac{\pi}{4} - 	an 	heta}{1 + 	an \frac{\pi}{4} 	an 	heta}$Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$:$= 	an \left(\frac{\pi}{4} - 	hetaight)$

$\sec 2 \theta - \tan 2 \theta =$

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