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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that $\cot x = \frac{\cos x}{\sin x}$. 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 identities $\cos ^2 x - \sin ^2 x = \cos 2x$ and $\cos ^2 x + \sin ^2 x = 1$: $= \frac{\cos(2 	imes 15^{\circ})}{1} = \cos 30^{\circ}$ 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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