Question

If tan theta equals a/b then (a sin theta -b cos theta )/(a sin theta + b cos theta) =(a2 - b2)/(a2+b2)

Answer 1

tan θ = a/b

sinθ/cosθ = a/b------(1)

lhs = (a sinθ- b cos θ)/(a sinθ + b cosθ)

= (asinθ/cosθ - bcosθ/cosθ)/(asinθ/cosθ+ bcosθ/cosθ)

dividing each term with cosθ

=(asinθ/cosθ-b/1)/(asinθ/cosθ+b/1)

=(a*a/b-b/1)/(a*a/b +b/1)
[from (1)]

= (a^2-b^2)/(a^2+b^2)
=rhs

Answer 2

tantheta = a/b ,

AC=√a2+b2.

L.H.S=

asintheta-bcostheta/asintheta+bcostheta= a2-b2/a2+b2.

L.H.S.=

a(a/√a2+be) - b(b/√a2+b2)/a(a/√a2+be) + b(b/√a2+b2)

=a2 - b2/√a2 + b2/a2 + b2/√a2 + b2

=(√a2 + b2) will get canceled in numerator and denominator.

=a2-b2 /a2+b2

Hence Proved.