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

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Question

(B) We are given the expression: $\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}}$Substitute $\cot 15^\circ = \frac{{\cos 15^\circ}}{{\sin 15^\cir

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We are given the expression: $\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}}$Substitute $\cot 15^\circ = \frac{{\cos 15^\circ}}{{\sin 15^\circ}}$:$= \frac{{\frac{{\cos^2 15^\circ}}{{\sin^2 15^\circ}} - 1}}{{\frac{{\cos^2 15^\circ}}{{\sin^2 15^\circ}} + 1}}$Simplify the numerator and denominator by taking the common denominator $\sin^2 15^\circ$:$= \frac{{\cos^2 15^\circ - \sin^2 15^\circ}}{{\cos^2 15^\circ + \sin^2 15^\circ}}$Using the trigonometric identities $\cos^2 	heta - \sin^2 	heta = \cos(2	heta)$ and $\cos^2 	heta + \sin^2 	heta = 1$:$= \cos(2 	imes 15^\circ) = \cos 30^\circ$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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