tan 75^circ - cot 75^circ = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that $	an 75^\circ = 	an(45^\circ + 30^\circ) = \frac{	an 45^\circ + 	an 30^\circ}{1 - 	an 45^\circ 	an 30^\circ} = \frac{1 + 1/\sqr

Vedclass · Accepted Answer

$2\sqrt{3}$

Vedclass · Answer

(A) We know that $	an 75^\circ = 	an(45^\circ + 30^\circ) = \frac{	an 45^\circ + 	an 30^\circ}{1 - 	an 45^\circ 	an 30^\circ} = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 1 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Since $\cot 75^\circ = \frac{1}{	an 75^\circ} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.Therefore,$	an 75^\circ - \cot 75^\circ = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}$.

$\tan 75^\circ - \cot 75^\circ = $

Similar Questions

If $A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$,then $\tan A \cdot \tan B+\tan B \cdot \tan C+\tan C \cdot \tan A$ is equal to

$\frac{1}{4} [\sqrt{3} \cos 23^\circ - \sin 23^\circ] = $

If $\tan(A + B) = p$ and $\tan(A - B) = q,$ then the value of $\tan(2A)$ in terms of $p$ and $q$ is

If $\cos (\theta+\phi)=\frac{3}{5}$ and $\sin (\theta-\phi)=\frac{5}{13}$, where $0 < \theta, \phi < \frac{\pi}{4}$, then $\cot (2 \theta)$ has the value:

If $\cos 43^{\circ} + \sin 43^{\circ} = k$, then find the value of $\cos 2^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module