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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Solve the equation $\sec^22x=1-\tan 2x$

$\begin{array}{1 1}(A)\;\large\frac{n\pi}{2}- \frac{\pi}{8}&(B)\;\large\frac{n\pi}{4}- \frac{\pi}{6}\\(C)\;\large\frac{n\pi}{3}- \frac{\pi}{12}&(D)\;\large\frac{n\pi}{6}- \frac{\pi}{12}\end{array} $

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1 Answer

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Toolbox:
  • $\sec^2\theta=\tan^2\theta+1$
$\sec^22x=1-\tan 2x$
$1+\tan^22x=1-\tan 2x$
$\tan^22x+\tan 2x=0$
$\tan 2x(1+\tan 2x)=0$
$\tan 2x=0$ or $\tan 2x+1=0$
$\Rightarrow \tan 2x=0$ or $\tan 2x=-1$
Now $\tan 2x=0\Rightarrow 2x=n\pi$
$x=\large\frac{n\pi}{2}$$,n\in I$
Again $\tan 2x=-1\Rightarrow \tan 2x=\tan(-\large\frac{\pi}{4})$
$2x=n\pi-\large\frac{\pi}{4}$
$x=\large\frac{n\pi}{2}-\frac{\pi}{8}$$\quad n\in I$
Hence (A) is the correct answer.
answered May 22, 2014 by sreemathi.v
 

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