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

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

Question

(B) Given equations are:$a \cos 	heta + b \sin 	heta = p$ ....$(1)$$a \sin 	heta - b \cos 	heta = q$ ....$(2)$Squaring both equations:$(a \cos 	h

Vedclass · Accepted Answer

$a^{2} + b^{2} = p^{2} + q^{2}$

Vedclass · Answer

(B) Given equations are:$a \cos 	heta + b \sin 	heta = p$ ....$(1)$$a \sin 	heta - b \cos 	heta = q$ ....$(2)$Squaring both equations:$(a \cos 	heta + b \sin 	heta)^{2} = p^{2}$$a^{2} \cos^{2} 	heta + b^{2} \sin^{2} 	heta + 2ab \sin 	heta \cos 	heta = p^{2}$ ....$(3)$$(a \sin 	heta - b \cos 	heta)^{2} = q^{2}$$a^{2} \sin^{2} 	heta + b^{2} \cos^{2} 	heta - 2ab \sin 	heta \cos 	heta = q^{2}$ ....$(4)$Adding equation $(3)$ and $(4)$:$(a^{2} \cos^{2} 	heta + a^{2} \sin^{2} 	heta) + (b^{2} \sin^{2} 	heta + b^{2} \cos^{2} 	heta) + (2ab \sin 	heta \cos 	heta - 2ab \sin 	heta \cos 	heta) = p^{2} + q^{2}$$a^{2}(\cos^{2} 	heta + \sin^{2} 	heta) + b^{2}(\sin^{2} 	heta + \cos^{2} 	heta) = p^{2} + q^{2}$Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$a^{2} + b^{2} = p^{2} + q^{2}$

If $a \cos \theta + b \sin \theta = p$ and $a \sin \theta - b \cos \theta = q$,then the relation between $a, b, p,$ and $q$ is:

Similar Questions

If $\sin \alpha = \frac{8}{17}, 0 < \alpha < 90^{\circ}$ and $\tan \beta = \frac{5}{12}, 0 < \beta < 90^{\circ},$ then $\cos (\alpha - \beta)$ is

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \left( \frac{x + y}{2} \right) = $

$\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ} = $

Given $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$,then which of the following holds good?

If $\alpha$ is a root of $25\cos^2\theta + 5\cos\theta - 12 = 0$ and $\pi/2 < \alpha < \pi$,then $\sin 2\alpha$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module