If acos theta + bsin theta = m and asin theta | vedclass

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Question

(c) Given that $a\,\cos 	heta + b\,\sin 	heta = m$ 

and $a\,\sin \,	heta - b\,\cos 	heta = n.$ 

Squaring and adding, we get

${(

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Given that $a\,\cos 	heta + b\,\sin 	heta = m$ 

and $a\,\sin \,	heta - b\,\cos 	heta = n.$ 

Squaring and adding, we get

${(a\,\,\cos 	heta + b\,\sin 	heta )^2} + {(a\,\sin 	heta - b\,\cos 	heta )^2} = {m^2} + {n^2}$

$ \Rightarrow \,\,{a^2}({\cos ^2}	heta + {\sin ^2}	heta ) + {b^2}({\cos ^2}	heta + {\sin ^2}	heta )$ 

$ + 2ab\,(\cos 	heta \,\sin 	heta - \sin 	heta \,\cos 	heta ) = {m^2} + {n^2}$ Hence, ${a^2} + {b^2} = {m^2} + {n^2}.$

Trick : Here we can guess that the value of ${a^2} + {b^2}$ is independent of $	heta$, so put any suitable value of $	heta$

$i.e.$ $\frac{\pi }{2},$ so that $b = m$ and $a = n.$ 

Hence ${a^2} + {b^2} = {m^2} + {n^2}.$

(Also check for other value of $	heta$).

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

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