Find the value of operatorname{cosec} 750^{ci | vedclass

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

Question

(A) We have the expression $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.Using the periodicity of trigonometric functions,$\operatorname{cos

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We have the expression $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.Using the periodicity of trigonometric functions,$\operatorname{cosec}(n 	imes 360^{\circ} + 	heta) = \operatorname{cosec} 	heta$ and $\cot(n 	imes 360^{\circ} + 	heta) = \cot 	heta$.$\operatorname{cosec} 750^{\circ} = \operatorname{cosec}(2 	imes 360^{\circ} + 30^{\circ}) = \operatorname{cosec} 30^{\circ} = 2$.$\cot 765^{\circ} = \cot(2 	imes 360^{\circ} + 45^{\circ}) = \cot 45^{\circ} = 1$.Substituting these values into the expression:$2 - 2(1) = 2 - 2 = 0$.Thus,the correct option is $A$.

Find the value of $\operatorname{cosec} 750^{\circ} - 2 \cot 765^{\circ}$.

Similar Questions

If $\theta$ lies in the first quadrant and $5 \tan \theta = 4$,then $\frac{5 \sin \theta - 3 \cos \theta}{\sin \theta + 2 \cos \theta}$ is equal to

If $\sec x + \tan x = 3, x \in (0, \frac{\pi}{2})$,then $\sin x =$

The value of $\tan \frac{\pi}{3} + 2 \tan \frac{2 \pi}{3} + 4 \tan \frac{4 \pi}{3} + 8 \tan \frac{8 \pi}{3}$ is equal to (in $\sqrt{3}$)

$\frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$ is equal to

If $5 \tan \theta = 4$,then $\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module