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3.Trigonometrical Ratios, Functions and Identities
easy
If $\tan \theta = - \frac{1}{{\sqrt {10} }}$ and $\theta $ lies in the fourth quadrant, then $\cos \theta = $
A
$1/\sqrt {11} $
B
$ - 1/\sqrt {11} $
C
$\sqrt {\frac{{10}}{{11}}} $
D
$ - \sqrt {\frac{{10}}{{11}}} $
Solution
(c) We have $\tan \theta = – \frac{1}{{\sqrt {10} }},$
therefore $\theta $ is in $IV$ quadrant.
So $\cos \theta = + ve$.Now $1 + {\tan ^2}\theta = {\sec ^2}\theta $
$\Rightarrow 1 + \frac{1}{{10}} = {\sec ^2}\theta $
$ \Rightarrow {\sec ^2}\theta = \frac{{11}}{{10}}$
$\Rightarrow \cos \theta = \sqrt {\left( {\frac{{10}}{{11}}} \right)} $.
Standard 11
Mathematics
