If $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ then $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ equals :-
If $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ then $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ equals :-
- A
$\frac{{\sqrt {(1 - {a^2})} - a\sqrt 3 }}{2}$
- B
$\frac{{1 - {a^2} + a\sqrt 3 }}{2}$
- C
$\frac{{a\sqrt 3 - \sqrt {(1 - {a^2})} }}{2}$
- D
$\frac{{ - \sqrt {(1 - {a^2})} - a\sqrt 3 }}{2}$
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If $\theta = 3\, \alpha$ and $sin\, \theta =$ $\frac{a}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$. The value of the expression , $a \,cosec\, \alpha - b \,sec\, \alpha$ is
If $\theta = 3\, \alpha$ and $sin\, \theta =$ $\frac{a}{{\sqrt {{a^2}\,\, + \,\,{b^2}} }}$. The value of the expression , $a \,cosec\, \alpha - b \,sec\, \alpha$ is
Show that
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
Show that
$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
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If $\sin \theta+\cos \theta=\frac{1}{2}$, then $16(\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta))$ is equal to:
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- [JEE MAIN 2021]