$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$ is equal to
$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$ is equal to
- A
$cot9^o$
- B
$tan9^o$
- C
$cot54^o$
- D
$tan54^o$
Similar Questions
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
- [IIT 1964]
If $a{\sin ^2}x + b{\cos ^2}x = c,\,\,$$b\,{\sin ^2}y + a\,{\cos ^2}y = d$ and $a\,\tan x = b\,\tan y,$ then $\frac{{{a^2}}}{{{b^2}}}$ is equal to
If $a{\sin ^2}x + b{\cos ^2}x = c,\,\,$$b\,{\sin ^2}y + a\,{\cos ^2}y = d$ and $a\,\tan x = b\,\tan y,$ then $\frac{{{a^2}}}{{{b^2}}}$ is equal to
$\sin {163^o}\cos {347^o} + \sin {73^o}\sin {167^o} = $
$\sin {163^o}\cos {347^o} + \sin {73^o}\sin {167^o} = $
If $\tan x = \frac{b}{a},$ then $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
If $\tan x = \frac{b}{a},$ then $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
The value of ,$\sqrt 3 \, cosec\, 20^o - sec\, 20^o $ is :
The value of ,$\sqrt 3 \, cosec\, 20^o - sec\, 20^o $ is :