If (sin theta - operatorname{cosec} theta)^2 | vedclass

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

Question

(A) Given: $(\sin 	heta - \operatorname{cosec} 	heta)^2 + (\cos 	heta + \sec 	heta)^2 = 5$ Expanding the squares: $(\sin^2 	heta + \operatorname{

Vedclass · Accepted Answer

$-2 \sqrt{2}$

Vedclass · Answer

(A) Given: $(\sin 	heta - \operatorname{cosec} 	heta)^2 + (\cos 	heta + \sec 	heta)^2 = 5$ Expanding the squares: $(\sin^2 	heta + \operatorname{cosec}^2 	heta - 2) + (\cos^2 	heta + \sec^2 	heta + 2) = 5$ Using $(\sin^2 	heta + \cos^2 	heta) = 1$: $1 + \operatorname{cosec}^2 	heta + \sec^2 	heta = 5$ Substitute $\operatorname{cosec}^2 	heta = 1 + \cot^2 	heta$ and $\sec^2 	heta = 1 + 	an^2 	heta$: $1 + (1 + \cot^2 	heta) + (1 + 	an^2 	heta) = 5$ $3 + \cot^2 	heta + 	an^2 	heta = 5 \Rightarrow 	an^2 	heta + \frac{1}{	an^2 	heta} = 2$ Let $x = 	an^2 	heta$,then $x + \frac{1}{x} = 2$ $\Rightarrow x^2 - 2x + 1 = 0$ $\Rightarrow (x - 1)^2 = 0$ $\Rightarrow 	an^2 	heta = 1$ Since $	heta$ is in the third quadrant,$	an 	heta = 1$ implies $	heta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$ Then $\sin 	heta = -\frac{1}{\sqrt{2}}$ and $\cos 	heta = -\frac{1}{\sqrt{2}}$ Therefore,$(\sin 	heta + \cos 	heta)^3 = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})^3 = (-\frac{2}{\sqrt{2}})^3 = (-\sqrt{2})^3 = -2\sqrt{2}$

If $(\sin \theta - \operatorname{cosec} \theta)^2 + (\cos \theta + \sec \theta)^2 = 5$ and $\theta$ lies in the third quadrant,then $(\sin \theta + \cos \theta)^3 = $

Similar Questions

If $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$,then $\sin 3A+\sin 3B$ is equal to

The geometric mean of $\tan 1^{\circ}, \tan 2^{\circ}, \ldots, \tan 89^{\circ}$ is

If a regular pentagon and a regular decagon have the same perimeter,then the ratio of their areas is

The value of $\text{cosec}10^{\circ} - \sqrt{3} \text{sec}10^{\circ}$ is equal to:

The value of $\cot 70^{\circ} + 4 \cos 70^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module