If tanθ = a/b then (asinθ - bcosθ)/(asinθ + bcosθ) ​

Question

If tanθ = a/b then \(\cfrac{(asinθ-bcosθ)}{(asinθ+bcosθ)}\) = ?

(asinθ - bcosθ)/(asinθ + bcosθ)

(a) \(\frac{(a^2+b^2)}{(a^2-b^2)}\)

(a² + b²)/(a² - b²)

(b) \(\frac{(a^2-b^2)}{(a^2+b^2)}\)

(a² + b²)/(a² + b²)

(c) \(\frac{a^2}{(a^2+b^2)}\)

a²/(a² + b²)

(d) \(\frac{a^2}{(a^2-b^2)}\)

a²/(a² - b²)

Answer 1

Correct answer is(b) \(\cfrac{(a^2-b^2)}{(a^2+b^2)}\)

We have tanθ = a/b

now, Dividing the numerator and denominator of the given expression by cos θ, we get:

\(\cfrac{(asinθ-bcosθ)}{(asinθ+bcosθ)}\)