જો 90^circ

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

Question

(d) $\sin \,A = \frac{4}{5}$

==>$	an A = - \frac{4}{3}$, $({90^o} < A < {180^o})$

$	an A = \frac{{2	an \frac{A}{2}}}{{1 - {{	an }^2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(d) $\sin \,A = \frac{4}{5}$ ==>$ an A = - \frac{4}{3}$, $({90^o} < A < {180^o})$ $ an A = \frac{{2 an \frac{A}{2}}}{{1 - {{ an }^2}\frac{A}{2}}}$, (Let $ an \frac{A}{2} = P$) ==> $ - \frac{4}{3} = \frac{{2P}}{{1 - {P^2}}}$ ==> $4{P^2} - 6P - 4 = 0$ ==> $P = \frac{{ - 1}}{2}{ m{ (impossible),}}\,$ hence $ an \frac{A}{2} = 2$.

જો $90^\circ < A < 180^\circ $ અને $\sin A = \frac{4}{5},$ તો $\tan \frac{A}{2} = . . .$

Similar Questions

જો $x + \frac{1}{x} = 2\,\cos \theta ,$ તો ${x^3} + \frac{1}{{{x^3}}} = $

ત્રિકોણ $ABC$ માં , ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2} = . . . .$

જો ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ તો $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $

$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$=

$\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)$ ની કિંમત $\dots\dots$છે.