If tan theta = - frac{1}{{sqrt {10} }} and th | vedclass

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(c) We have $	an 	heta = - \frac{1}{{\sqrt {10} }},$

therefore $	heta $ is in $IV$ quadrant. 

So $\cos 	heta = + ve$.Now $1 + {	an ^2}

Vedclass · Accepted Answer

$\sqrt {\frac{{10}}{{11}}} $

Vedclass · Answer

(c) We have $	an 	heta = - \frac{1}{{\sqrt {10} }},$

therefore $	heta $ is in $IV$ quadrant. 

So $\cos 	heta = + ve$.Now $1 + {	an ^2}	heta = {\sec ^2}	heta $

$\Rightarrow 1 + \frac{1}{{10}} = {\sec ^2}	heta $

$ \Rightarrow {\sec ^2}	heta = \frac{{11}}{{10}}$

$\Rightarrow \cos 	heta = \sqrt {\left( {\frac{{10}}{{11}}} ight)} $.

If $\tan \theta = - \frac{1}{{\sqrt {10} }}$ and $\theta $ lies in the fourth quadrant, then $\cos \theta = $

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