frac{cos 13^{circ}-sin 13^{circ}}{cos 13^{cir | vedclass

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

Question

(C) Given expression: $\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$Divide the numerator and den

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given expression: $\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$Divide the numerator and denominator of the first term by $\cos 13^{\circ}$:$= \frac{1-	an 13^{\circ}}{1+	an 13^{\circ}} + 	an 148^{\circ}$Using the formula $	an(A-B) = \frac{	an A - 	an B}{1+	an A 	an B}$,where $A=45^{\circ}$ and $B=13^{\circ}$:$= 	an(45^{\circ}-13^{\circ}) + 	an(180^{\circ}-32^{\circ})$$= 	an 32^{\circ} - 	an 32^{\circ} = 0$

$\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$ is equal to

Similar Questions

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

If $\tan 20^{\circ}=\lambda$,then the value of $\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\left(\tan 160^{\circ}\right)\left(\tan 110^{\circ}\right)}$ is:

If $\cos^3 80^{\circ} + \cos^3 40^{\circ} - \cos^3 20^{\circ} = k$,then $\frac{4k}{3} =$

$\frac{\cos 15^{\circ} \cos^2 22\frac{1}{2}^{\circ} - \sin 75^{\circ} \sin^2 52\frac{1}{2}^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}} = $

If $\sin \theta + \cos \theta = 1$,then $\sin \theta \cos \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module