3.Trigonometrical Ratios, Functions and Identities
easy

If $\theta $ lies in the second quadrant, then the value of $\sqrt {\left( {\frac{{1 - \sin \theta }}{{1 + \sin \theta }}} \right)} + \sqrt {\left( {\frac{{1 + \sin \theta }}{{1 - \sin \theta }}} \right)} $

A

$2\sec \theta $

B

$ - 2\sec \theta $

C

$2{\rm{cosec}} \, \theta $

D

None of these

Solution

(b) $\sqrt {\left( {\frac{{1 – \sin \theta }}{{1 + \sin \theta }}} \right)} + \sqrt {\left( {\frac{{1 + \sin \theta }}{{1 – \sin \theta }}} \right)} $

is the sum of two positive quantities and hence the result must be positive. 

But for $\frac{\pi }{2} < \theta < \pi ,$

we have the sum equal to $\frac{{1 – \sin \theta + 1 + \sin \theta }}{{\sqrt {1 – {{\sin }^2}\theta } }} = \frac{2}{{\cos \theta }};$ 

which is negative.

( $\because$ $\cos \theta $ is negative for $\theta$ lying in $2^{nd}$ quadrant). 

So the required positive value $ = \frac{{ – 2}}{{\cos \theta }} = – 2\,\sec \theta ,\,\left( {\frac{\pi }{2} < \theta < \pi } \right)$.

Standard 11
Mathematics

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