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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

The value of $\large\frac{1-\tan^215}{1+\tan^215}$ is

$\begin{array}{1 1}(A)\;1&(B)\;\sqrt 3\\(C)\;\large\frac{\sqrt 3}{2}&(D)\;2\end{array} $

1 Answer

Toolbox:
  • $\tan \theta=\large\frac{\sin\theta}{\cos\theta}$
  • $\sin^2\theta+\cos^2\theta=1$
  • $\cos^2\theta-\sin^2\theta=\cos 2\theta$
Given :
$\large\frac{1-\tan^215}{1+\tan^215}$
$\large\frac{1-\large\frac{\sin^215}{\cos^215}}{1+\large\frac{\sin^215}{\cos^215}}$
$\Rightarrow \large\frac{\Large\frac{\cos^215-\sin^215}{\cos^215}}{\Large\frac{\cos^215+\sin^215}{\cos^215}}$
$\Rightarrow \cos^15-\sin^215$
$\Rightarrow \cos (2\times 15)^{\large\circ}=\cos 30^{\large\circ}=\large\frac{\sqrt 3}{2}$
Hence (C) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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