If tan θ = \(\frac{a}{b}\), prove that \(\frac {a sin\, θ - b cos\,

1.	If tan θ = \(\frac{a}{b}\), prove that \(\frac {a sin\, θ - b cos\, θ}{a sin\, θ + b cos\, θ}\) = \(\frac{a^2 - b^2}{a^2 + b^2}\).
Answer» Given, tan θ = \(\frac{a}{b}\) \(\frac {a sin\, θ - b cos\, θ}{a sin\, θ + b cos\, θ}\) From LHS, let’s divide the numerator and denominator by cos θ. And we get, \(\frac{(a\, tan\, θ\, –\, b)}{(a\, tan\, θ\, +\, b\,) }\) ⇒ \(\frac{(a(a/b) – b)}{(a(a/b) + b)}\) [using the value of tan θ] ⇒ \(\frac{(a^2 – b^2)}{b^2} \over \frac{(a^2 + b^2)}{b^2}\) [After taking LCM and simplifying it] ⇒ \(\frac{(a^2 – b^2)}{(a^2 + b^2)}\) = RHS Hence Proved

If tan θ = \(\frac{a}{b}\), prove that \(\frac {a sin\, θ - b cos\, θ}{a sin\, θ + b cos\, θ}\) = \(\frac{a^2 - b^2}{a^2 + b^2}\).

Answer»

Given, tan θ = \(\frac{a}{b}\)

\(\frac {a sin\, θ - b cos\, θ}{a sin\, θ + b cos\, θ}\)

From LHS, let’s divide the numerator and denominator by cos θ.

And we get,

\(\frac{(a\, tan\, θ\, –\, b)}{(a\, tan\, θ\, +\, b\,) }\)

⇒ \(\frac{(a(a/b) – b)}{(a(a/b) + b)}\) [using the value of tan θ]

⇒ \(\frac{(a^2 – b^2)}{b^2} \over \frac{(a^2 + b^2)}{b^2}\) [After taking LCM and simplifying it]

⇒ \(\frac{(a^2 – b^2)}{(a^2 + b^2)}\) = RHS

Hence Proved