frac{cos 9^circ + sin 9^circ}{cos 9^circ - si | vedclass

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

Question

(A) Divide the numerator and denominator by $\cos 9^\circ$:$\frac{\frac{\cos 9^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 9^\circ}}{\frac{\cos 9^

Vedclass · Accepted Answer

$	an 54^\circ$

Vedclass · Answer

(A) Divide the numerator and denominator by $\cos 9^\circ$:$\frac{\frac{\cos 9^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 9^\circ}}{\frac{\cos 9^\circ}{\cos 9^\circ} - \frac{\sin 9^\circ}{\cos 9^\circ}} = \frac{1 + 	an 9^\circ}{1 - 	an 9^\circ}$Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = 45^\circ$ and $B = 9^\circ$:$\frac{	an 45^\circ + 	an 9^\circ}{1 - 	an 45^\circ 	an 9^\circ} = 	an(45^\circ + 9^\circ) = 	an 54^\circ$.

$\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = $

Similar Questions

Prove that $\frac{\tan (\frac{\pi}{4}+x)}{\tan (\frac{\pi}{4}-x)} = (\frac{1+\tan x}{1-\tan x})^{2}$.

$\frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} = $

If $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$,then $\alpha + \beta = $

$\frac{\sin(B + A) + \cos(B - A)}{\sin(B - A) + \cos(B + A)} = $

In an isosceles right-angled triangle,a straight line is drawn from the midpoint of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

Vedclass Test Series

Exam Paper Generator

Online Exam Module