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

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Answer 1 · 2017-02-11T15:16:26

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 · 2017-02-11T15:47:53

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

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 · 2017-02-12T18:04:57

#(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