If $\tan A = \frac{1}{2},$ then $\tan 3A = $
If $\tan A = \frac{1}{2},$ then $\tan 3A = $
- A
$\frac{9}{2}$
- B
$\frac{{11}}{2}$
- C
$\frac{7}{2}$
- D
$ - \frac{1}{2}$
Similar Questions
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$
The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :
The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :
If $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ then $\cos 2\alpha = $
If $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ then $\cos 2\alpha = $
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
- [IIT 1980]
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$