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  4. The value of (1-tan2 15°/1+tan2 15°) is

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

Trigonometric Functions

Solution:

$\frac{1-tan^{2}\,15^{\circ}}{1+tan^{2}\,15^{\circ}}=\frac{cos^{2}\,15^{\circ}-sin^{2}\,15^{\circ}}{cos^{2}\,15^{\circ}+sin^{2}\,15^{\circ}}$
$=\frac{cos\left(2 \times 15^{\circ}\right)}{1}$
$=cos\,30^{\circ}=\frac{\sqrt{3}}{2}$