If a cos theta + b sin theta = m and a sin th | vedclass

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

Question

(C) Given that $a \cos 	heta + b \sin 	heta = m$ $(1)$ and $a \sin 	heta - b \cos 	heta = n$ $(2)$ Squaring both equations and adding them,we get:

Vedclass · Accepted Answer

${m^2} + {n^2}$

Vedclass · Answer

(C) Given that $a \cos 	heta + b \sin 	heta = m$ $(1)$ and $a \sin 	heta - b \cos 	heta = n$ $(2)$ Squaring both equations and adding them,we get: $(a \cos 	heta + b \sin 	heta)^2 + (a \sin 	heta - b \cos 	heta)^2 = m^2 + n^2$ Expanding the squares: $(a^2 \cos^2 	heta + b^2 \sin^2 	heta + 2ab \sin 	heta \cos 	heta) + (a^2 \sin^2 	heta + b^2 \cos^2 	heta - 2ab \sin 	heta \cos 	heta) = m^2 + n^2$ Grouping the terms: $a^2(\cos^2 	heta + \sin^2 	heta) + b^2(\sin^2 	heta + \cos^2 	heta) = m^2 + n^2$ Since $\sin^2 	heta + \cos^2 	heta = 1$,we have: $a^2(1) + b^2(1) = m^2 + n^2$ Therefore,$a^2 + b^2 = m^2 + n^2$.

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

Similar Questions

If $\frac{1 - \cos x}{\cos x(1 + \cos x)} = \frac{\sin \alpha}{\cos x} - \frac{2}{1 + \cos x}$,then $\alpha = $

The value of $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right)$ is equal to

If $x = 3 \sin \theta$,$y = 3 \cos \theta \cos \phi$,and $z = 3 \cos \theta \sin \phi$,then $x^{2} + y^{2} + z^{2} =$

$\sin ^4 \frac{\pi}{8}+\cos ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\cos ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=$

If $\cos \theta, \sin \theta$ and $\cot \theta$ are in geometric progression,then $\sin ^9 \theta+\sin ^6 \theta+3 \sin ^5 \theta+\sin ^3 \theta+\sin ^2 \theta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module