tan 9^circ - tan 27^circ - tan 63^circ + tan | vedclass

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

Question

(c) $	an \,\,{9^o} - 	an \,\,{27^o} - 	an \,\,{63^o} + 	an \,\,{81^o}$

$ = 	an \,\,{9^o} - 	an \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c) $	an \,\,{9^o} - 	an \,\,{27^o} - 	an \,\,{63^o} + 	an \,\,{81^o}$

$ = 	an \,\,{9^o} - 	an \,\,{27^o} - \cot \,\,{27^o} + \cot \,\,{9^o}$

$ = (	an \,\,{9^o} + \cot \,\,{9^o}) - (	an \,\,{27^o} + \cot \,\,{27^o})$

$ = \frac{{\cos ({9^o} - {9^o})}}{{\sin {9^o}\cos {9^o}}} - \frac{{\cos ({{27}^o} - {{27}^o})}}{{\sin {{27}^o}.\cos {{27}^o}}} = \frac{2}{{\sin {{18}^o}}} - \frac{2}{{\sin {{54}^o}}}$

$ = 2\,\left\{ {\frac{{\sin \,\,{{54}^o} - \sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}\,\sin \,\,{{54}^o}}}} ight\} $

$= 2.\,\,\frac{{2\,.\,\cos \,\,{{36}^o}.\,\sin \,\,{{18}^o}}}{{\sin \,\,{{18}^o}.\,\sin \,\,{{54}^o}}} = 4$

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

Similar Questions

If $\cos A = \frac{3}{4}$, then $32\sin \frac{A}{2}\cos \frac{5}{2}A = $

$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $

If $\tan x + \tan \left( {\frac{\pi }{3} + x} \right) + \tan \left( {\frac{{2\pi }}{3} + x} \right) = 3,$ then

$\sqrt {2 + \sqrt {2 + 2\cos 4\theta } } = $

If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is