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}$.Rationalizing the denominator: $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.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

The value of ${e^{{{\log }_{10}}\tan 1^\circ + {{\log }_{10}}\tan 2^\circ + {{\log }_{10}}\tan 3^\circ + ........... + {{\log }_{10}}\tan 89^\circ }}$ is

If $0^{\circ} < \theta < 90^{\circ},$ the value of $\sin \theta + \cos \theta$ is

If $\tan x = \frac{2b}{a - c}$ $(a \ne c)$,$y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$,then:

If $a \cos^3 \alpha + 3a \cos \alpha \sin^2 \alpha = m$ and $a \sin^3 \alpha + 3a \cos^2 \alpha \sin \alpha = n$,then $(m + n)^{2/3} + (m - n)^{2/3}$ is equal to

If $A$ and $B$ are complementary angles,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module