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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the most general value of $\theta$ satisfying the equation $\tan\theta=-1$ and $\cos \theta=\large\frac{1}{\sqrt 2}$

$\begin{array}{1 1}(A)\;n\pi+\large\frac{9\pi}{4}&(B)\;2n\pi+\large\frac{7\pi}{4}\\(C)\;3n\pi+\large\frac{7\pi}{4}&(D)\;2n\pi+\large\frac{9\pi}{5}\end{array} $

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1 Answer

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$\tan \theta=-1=\tan\large\frac{3\pi}{4}$$=\tan\large\frac{7\pi}{4}$
$\tan(\pi-\large\frac{3\pi}{4})=$$\tan(\pi+\large\frac{3\pi}{4})$
$\cos\theta=\large\frac{1}{\sqrt 2}$
$\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$
$\cos(2\pi-\large\frac{\pi}{4})=\large\frac{1}{\sqrt 2}=$$\cos\large\frac{7\pi}{4}$
Principal value for both $\tan\theta$ and $\cos\theta$ is $\large\frac{7\pi}{4}$
$\therefore$ general value is $2n\pi+\large\frac{7\pi}{4}$
Hence (B) is the correct answer.
answered Apr 18, 2014 by sreemathi.v
 

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