If tanθ = a/b, prove that asinθ - bcosθ/asinθ + bcosθ = a^2 - b^2/a^2 + b^2 ​

Question

If tanθ = \(\frac{a}b\), prove that \(\frac{asinθ-bcosθ}{asinsθ+bcosθ}\) = \(\frac{a^2-b^2}{a^2+b^2}\)

Answer 1

Given: tan θ = \(\frac{a}b\)

To prove: \(\frac{asinθ-bcosθ}{asinsθ+bcosθ}\) = \(\frac{a^2-b^2}{a^2+b^2}\)..... (1) 

Solution: Consider LHS of eq. (1)Take b cosθ common from both numerator and denominator.

Put the value of tanθ to get,

hence proved