frac{{cot^2 15^circ - 1}}{{cot^2 15^circ + 1} | vedclass

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Question

(B) We know that $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$.Substituting this into the expression:$\frac{\cot^2 15^\circ - 1}{\cot^2 15^\circ + 1

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(B) We know that $\cot 	heta = \frac{\cos 	heta}{\sin 	heta}$.Substituting this into the expression:$\frac{\cot^2 15^\circ - 1}{\cot^2 15^\circ + 1} = \frac{\frac{\cos^2 15^\circ}{\sin^2 15^\circ} - 1}{\frac{\cos^2 15^\circ}{\sin^2 15^\circ} + 1}$$= \frac{\cos^2 15^\circ - \sin^2 15^\circ}{\cos^2 15^\circ + \sin^2 15^\circ}$Using the trigonometric identities $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$ and $\cos^2 	heta + \sin^2 	heta = 1$:$= \frac{\cos(2 	imes 15^\circ)}{1} = \cos 30^\circ$Since $\cos 30^\circ = \frac{\sqrt{3}}{2}$,the final answer is $\frac{\sqrt{3}}{2}$.

$\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}} = $

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