સાબિત કરો કે, $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
સાબિત કરો કે, $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
We know that $3 x=2 x+x$
Therefore, $\tan 3 x=\tan (2 x+x)$
or $\tan 3 x=\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}$
or $\tan 3 x-\tan 3 x \tan 2 x \tan x=\tan 2 x+\tan x$
or $\tan 3 x-\tan 2 x-\tan x=\tan 3 x \tan 2 x \tan x$
or $\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
Similar Questions
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
- [IIT 1974]
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
- [IIT 1964]
જો $\alpha ,\,\beta ,\,\gamma \in \,\left( {0,\,\frac{\pi }{2}} \right)$, તો $\frac{{\sin \,(\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }} = . . ..$
જો $\alpha ,\,\beta ,\,\gamma \in \,\left( {0,\,\frac{\pi }{2}} \right)$, તો $\frac{{\sin \,(\alpha + \beta + \gamma )}}{{\sin \alpha + \sin \beta + \sin \gamma }} = . . ..$
$cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ =
$cot\, 7\frac{{{1^0}}}{2}$ $+ tan\, 67 \frac{{{1^0}}}{2} - cot 67 \frac{{{1^0}}}{2} - tan7 \frac{{{1^0}}}{2}$ =
ત્રિકોણ $ABC$ માં , $\tan A + \tan B + \tan C = 6$ અને $\tan A\tan B = 2,$ તો $\tan A,\,\,\tan B$ અને $\tan C$ મેળવો.
ત્રિકોણ $ABC$ માં , $\tan A + \tan B + \tan C = 6$ અને $\tan A\tan B = 2,$ તો $\tan A,\,\,\tan B$ અને $\tan C$ મેળવો.