$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$ 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
$4 \,\,sin5^o \,\,sin55^o \,\,sin65^o$ has the values equal to
$4 \,\,sin5^o \,\,sin55^o \,\,sin65^o$ has the values equal to
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
- [IIT 1988]
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}}} = $
If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to
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If $\cos A = \frac{3}{4}$, then $32\sin \frac{A}{2}\cos \frac{5}{2}A = $
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