$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ is equal to
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ is equal to
- A
$0$
- B
$4\tan {40^o}$
- C
$8\tan {40^o}$
- D
$8\cot {40^o}$
Similar Questions
If $\theta $ is an acute angle and $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, then $\tan \theta $ is equal to
If $\theta $ is an acute angle and $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, then $\tan \theta $ is equal to
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
If $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ then $x = $
If $\cos A = \cos B\,\,\cos C$and $A + B + C = \pi ,$ then the value of $\cot \,B\,\cot \,C$ is
If $\cos A = \cos B\,\,\cos C$and $A + B + C = \pi ,$ then the value of $\cot \,B\,\cot \,C$ is
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
If $a\tan \theta = b$, then $a\cos 2\theta + b\sin 2\theta = $
If $a\tan \theta = b$, then $a\cos 2\theta + b\sin 2\theta = $