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Question
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
Solution
We have `m^2 + n^2 = [(a cos theta + b sin theta)^2 + ( a sin theta - b cos theta )^2 ]`
=` ( a^2 cos^2 theta + b^2 sin ^2 theta + 2 ab cos theta sin theta)`
+`(a^2 sin^2 theta + b^2 cos^2 theta -2ab cos theta sin theta)`
=`a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 vos^2 theta`
=`(a^2 cos^2 theta + b^2 sin^2 theta) + ( b^2 cos^2 theta + b^2 sin^2 theta )`
=`a^2 (cos^2 theta + sin^2 theta ) + b^2 ( cos^2 theta + sin^2 theta )`
=`a^2 + b^2 [∵ sin^2 + cos^2 = 1]`
Hence , `m^2 + n^2 = a^2 + b^2`
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