The value of tan 9^{circ} - tan 27^{circ} - t | vedclass

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

Question

(C) Given expression is $E = 	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$.Using $	an 81^{\circ} = \cot 9^{\circ}$ and $	a

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given expression is $E = 	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$.Using $	an 81^{\circ} = \cot 9^{\circ}$ and $	an 63^{\circ} = \cot 27^{\circ}$,we get:$E = (\cot 9^{\circ} + 	an 9^{\circ}) - (\cot 27^{\circ} + 	an 27^{\circ})$.Since $\cot 	heta + 	an 	heta = \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$,we have:$E = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.Using values $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$,we get:$E = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = \frac{8(\sqrt{5}+1) - 8(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{8\sqrt{5} + 8 - 8\sqrt{5} + 8}{5-1} = \frac{16}{4} = 4$.

The value of $\tan 9^{\circ} - \tan 27^{\circ} - \tan 63^{\circ} + \tan 81^{\circ}$ is

Similar Questions

$\Delta XYZ$ is right-angled at $Y$. If $\cos X = \frac{3}{5}$,then what is the value of $\operatorname{cosec} Z$?

$\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \dots \cos 179^\circ = $

If $\cot \theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n,$ then which of the following is correct?

The value of $\frac{\tan 70^\circ - \tan 20^\circ}{\tan 50^\circ} = $

If $b \sin \alpha = a \sin (\alpha + 2\beta)$,then $\frac{a + b}{a - b} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module