If cos theta = frac{x^{2}-y^{2}}{x^{2}+y^{2}} | vedclass

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

Question

(D) Given $\cos 	heta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - \cos

Vedclass · Accepted Answer

$\frac{x^{2}-y^{2}}{2xy}$

Vedclass · Answer

(D) Given $\cos 	heta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.$\sin 	heta = \sqrt{1 - \left(\frac{x^2 - y^2}{x^2 + y^2}ight)^2} = \sqrt{\frac{(x^2 + y^2)^2 - (x^2 - y^2)^2}{(x^2 + y^2)^2}}$.Using $(a+b)^2 - (a-b)^2 = 4ab$,we get $\sin 	heta = \sqrt{\frac{4x^2y^2}{(x^2 + y^2)^2}} = \frac{2xy}{x^2 + y^2}$.Now,$\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{(x^2 - y^2) / (x^2 + y^2)}{(2xy) / (x^2 + y^2)}$.$\cot 	heta = \frac{x^2 - y^2}{2xy}$.

If $\cos \theta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$,then the value of $\cot \theta$ is equal to (where $0^{\circ} \leq \theta \leq 90^{\circ}$)

Similar Questions

$\log \sin 1^{\circ} \cdot \log \sin 2^{\circ} \cdot \log \sin 3^{\circ} \cdot \ldots \cdot \log \sin 179^{\circ} = ?$

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $ax^2 + bx + c = 0$ $(c \neq 0)$,then:

$\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}$ is equal to

If $\cos A = m \cos B,$ then

If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$ where $0 < \alpha, \beta < \frac{\pi}{2}$,then $2\beta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module