Question

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)`

Answer 1

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 𝜃, 𝑤𝑒 𝑔𝑒𝑡:

`(a tan theta-b)/(a tan theta +b)  (∴ tan theta = (sin theta)/(cos theta))`

Now, substituting the value of tan 𝜃 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡:

`(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.