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If $\sin x=-\large\frac{2\sqrt 6}{5}$ and $x$ lies in quadrant III,find the values of $\cos x$ and $\cot x$

1 Answer

  • $\cot^2\theta=cosec^2\theta-1$
  • $\cot \theta=\sqrt{cosec^2\theta-1}$
  • $\tan \theta=\large\frac{\sin \theta}{\cos \theta}$
  • $\cot \theta=\large\frac{\cos \theta}{\sin \theta}$
Clearly $x$ lies in the third quadrant in which $\tan\theta$ and $\cot\theta$ are positive and all the other trignometric functions are negative.
Now,$\sin x=-\large\frac{2\sqrt 6}{5}$
$cosec x=\large\frac{1}{\sin x}=-\frac{5}{2\sqrt 6}$
$\therefore \cot \theta=\sqrt{cosec^2\theta-1}$
$\therefore \cot \theta=\sqrt{\large\frac{25}{4\times 6}-1}$
$\Rightarrow \sqrt{\large\frac{25-24}{24}}$
$\Rightarrow \sqrt{\large\frac{1}{24}}$
$\Rightarrow \sqrt{\large\frac{1}{2\sqrt 6}}$
$\cot x=\large\frac{\cos x}{\sin x}$
$\cos x=\cot x.\sin x$
$\Rightarrow \large\frac{1}{2\sqrt 6}\times \frac{-2\sqrt 6}{5}$
$\Rightarrow -\large\frac{1}{5}$
$\cos x=-\large\frac{1}{5},$$\cot x=\large\frac{1}{2\sqrt 6}$
answered Apr 22, 2014 by sreemathi.v

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