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If tan `theta = a/b`, show that `((a sin theta - b cos theta))/((a sin theta + bcos theta))= ((a^2-b^2))/(a^2+b^2)`
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It is given that tan `theta = a/b`
LHS = `(a sin theta - b cos theta )/(a sin theta + b cos theta)`
Dividing the numerator and denominator by cos ЁЭЬГ, ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб:
`(a tan theta-b)/(a tan theta +b) (∴ tan theta = (sin theta)/(cos theta))`
Now, substituting the value of tan ЁЭЬГ ЁЭСЦЁЭСЫ ЁЭСбтДОЁЭСТ ЁЭСОЁЭСПЁЭСЬЁЭСгЁЭСТ ЁЭСТЁЭСеЁЭСЭЁЭСЯЁЭСТЁЭСаЁЭСаЁЭСЦЁЭСЬЁЭСЫ, ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб:
`(a(a/b)-b)/(a(a/b)+b)`
=`(a^2/b-b)/(a^2/b+b)`
= `(a^2 - b^2)/(a^2+b^2) = `RHS
i.e., LHS = RHS
Hence proved.
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