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

Find the general solution for each of the following equation $\sec^22x=1-\tan 2x$

$\begin{array}{1 1}(A)\;x=\large\frac{n\pi}{2}+\frac{3\pi}{8}&(B)\;x=\large\frac{n\pi}{3}+\frac{3\pi}{8}\\(C)\;x=\large\frac{n\pi}{3}+\frac{5\pi}{8}&(D)\;x=\large\frac{2n\pi}{5}+\frac{3\pi}{8}\end{array} $

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  • $\sec^22x=1+\tan ^22x$
$\sec^22x=1-\tan 2x$
$\Rightarrow 1+\tan^22x=1-\tan 2x=0$
$\Rightarrow \tan^22x+\tan 2x=0$
$\Rightarrow \tan 2x(\tan 2x+1)=0$
If $\tan 2x=0,2x=n\pi$ or $x=\large\frac{n\pi}{2}$
If $\tan 2x+1=0,\tan 2x=-1=\tan(\pi-\large\frac{\pi}{4})$$=\tan \large\frac{3\pi}{4}$
$\Rightarrow 2x=n\pi+\large\frac{3\pi}{4}$
Or $x=\large\frac{n\pi}{2}+\frac{3\pi}{8}$
Hence (A) is the correct answer.
answered Apr 18, 2014 by sreemathi.v
 

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