The value of $\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
The value of $\frac{{\tan {{70}^o} - \tan {{20}^o}}}{{\tan {{50}^o}}} = $
- A
$1$
- B
$2$
- C
$3$
- D
$0$
Similar Questions
Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to
Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$
- [IIT 1974]
$\tan \frac{A}{2}$ is equal to
$\tan \frac{A}{2}$ is equal to
If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for
If $\sin \theta + \cos \theta = x,$ then ${\sin ^6}\theta + {\cos ^6}\theta = \frac{1}{4}[4 - 3{({x^2} - 1)^2}]$ for
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :