If theta lies in the second quadrant, then t | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) $\sqrt {\left( {\frac{{1 - \sin 	heta }}{{1 + \sin 	heta }}} ight)} + \sqrt {\left( {\frac{{1 + \sin 	heta }}{{1 - \sin 	heta }}} ight)} $

Vedclass · Accepted Answer

$ - 2\sec 	heta $

Vedclass · Answer

(b) $\sqrt {\left( {\frac{{1 - \sin 	heta }}{{1 + \sin 	heta }}} ight)} + \sqrt {\left( {\frac{{1 + \sin 	heta }}{{1 - \sin 	heta }}} ight)} $

is the sum of two positive quantities and hence the result must be positive. 

But for $\frac{\pi }{2} < 	heta < \pi ,$

we have the sum equal to $\frac{{1 - \sin 	heta + 1 + \sin 	heta }}{{\sqrt {1 - {{\sin }^2}	heta } }} = \frac{2}{{\cos 	heta }};$ 

which is negative.

( $\because$ $\cos 	heta $ is negative for $	heta$ lying in $2^{nd}$ quadrant). 

So the required positive value $ = \frac{{ - 2}}{{\cos 	heta }} = - 2\,\sec 	heta ,\,\left( {\frac{\pi }{2} < 	heta < \pi } ight)$.

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)} $

Similar Questions

If $sin\theta_1 + sin\theta_2 + sin\theta_3 = 3,$ then $cos\theta_1 + cos\theta_2 + cos\theta_3=$

If $A = 130^\circ $ and $x = \sin A + \cos A,$ then

The equation ${\sin ^2}\theta = \frac{{{x^2} + {y^2}}}{{2xy}},x,y, \ne 0$ is possible if

If $\tan A + \cot A = 4,$ then ${\tan ^4}A + {\cot ^4}A$ is equal to

If $\sin A,\cos A$ and $\tan A$ are in $G.P.$, then ${\cos ^3}A + {\cos ^2}A$ is equal to