Question

If `tantheta = a/b`, then how do you show that `(asintheta - bcostheta)/(asintheta + bcostheta) = (a^2 - b^2)/(a^2 + b^2)`?

Answer 1

We know that `tantheta = sintheta/costheta`, so we know that `sintheta = a` and `costheta = b`

`(asintheta - bcostheta)/(asintheta + bcostheta) = (a(a) - b(b))/(a(a) + b(b))`

This can be written as `(a^2 - b^2)/(a^2 + b^2)`.

Hopefully this helps!

Answer 2

proved R.H.S = L.H.S

Explanation:

We know `tan theta = a/b =[height]/[base]`

In a right angle triangle, we know, `(hypotenuse)^2 = (height)^2 +(base)^2`

So, hypotenuse = `sqrt (a^2 + b^2)`

Now sin`theta = (height)/(hypotenuse) = a/[sqrt(a^2 +b^2)]` and
cos `theta = (base)/(hypotenuse) =b/[sqrt(a^2 + b^2)]`

So, `[a sin theta - b cos theta]/[a sin theta + b cos theta]`

= `[a.{a/sqrt(a^2+b^2)} - [b{b/sqrt(a^2+b^2)}]]/[[a{a/sqrt(a^2+b^2)}] +[b{b/sqrt(a^2+b^2)}]`

=`[a^2/sqrt(a^2+b^2)- b^2/sqrt(a^2+b^2)]/[a^2/sqrt(a^2+b^2)-b^2/sqrt(a^2+b^2)]`

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

= R.H.S

Answer 3

`(asintheta - bcostheta)/(asintheta + bcostheta)`

`=(asintheta/costheta - bcostheta/costheta)/(asintheta/costheta+ bcostheta/costheta)`

`=(atantheta - b)/(atantheta+ b)`

`=(axxa/b - b)/(axxa/b+ b)`

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

Proved