If operatorname{cosec} theta - sin theta = m | vedclass

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

Question

(B) Given $m = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $m = \operatorname{cosec} 	heta - \sin 	heta = \frac{1}{\sin 	heta} - \sin 	heta = \frac{1 - \sin^2 	heta}{\sin 	heta} = \frac{\cos^2 	heta}{\sin 	heta}$.Given $n = \sec 	heta - \cos 	heta = \frac{1}{\cos 	heta} - \cos 	heta = \frac{1 - \cos^2 	heta}{\cos 	heta} = \frac{\sin^2 	heta}{\cos 	heta}$.Now,consider the expression $(m^2 n)^{2/3} + (m n^2)^{2/3}$.Substitute the values of $m$ and $n$:$(m^2 n)^{2/3} = \left( \frac{\cos^4 	heta}{\sin^2 	heta} \cdot \frac{\sin^2 	heta}{\cos 	heta} ight)^{2/3} = (\cos^3 	heta)^{2/3} = \cos^2 	heta$.Similarly,$(m n^2)^{2/3} = \left( \frac{\cos^2 	heta}{\sin 	heta} \cdot \frac{\sin^4 	heta}{\cos^2 	heta} ight)^{2/3} = (\sin^3 	heta)^{2/3} = \sin^2 	heta$.Adding these results:$\cos^2 	heta + \sin^2 	heta = 1$.

If $\operatorname{cosec} \theta - \sin \theta = m$ and $\sec \theta - \cos \theta = n$,then $(m^2 n)^{2/3} + (m n^2)^{2/3} = $

Similar Questions

The value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$ is

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

If $\operatorname{cosec}^{2} \theta = \frac{4xy}{(x+y)^{2}},$ then

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $

If $A + B + C = \pi ,$ then $\cos \, 2A + \cos \, 2B + \cos \, 2C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module