The value of $\frac{1-\tan ^2 15^{\circ}}{1+\tan ^2 15^{\circ}}$ is

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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The value of

\frac{1-\tan ^2 15^{\circ}}{1+\tan ^2 15^{\circ}}

View answer and explanation

Correct option: C

\sqrt{3} / 2

Option C (

\sqrt{3} / 2

) is correct because the expression is a direct application of the cosine double angle formula:

\cos(2\theta) = \frac{{1 - \tan^2(\theta)}}{{1 + \tan^2(\theta)}}

with

\theta = 15^{\circ}

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\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=