Question

If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn

Answer 1

Let
`a=\tan A` and
`b=\sin A`. Then

`m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab`

`\implies m^2-n^2=4\tan A\sin A`

and

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

`\implies4√(mn)=4√(\tan^2A-\sin^2A)`

The expression under the square root can be rewritten as

`\tan^2A-\sin^2A=(\sin^2A)/(\cos^2A)-\sin^2A=\sin^2A\left(\frac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)`

Recall that

`\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A`

so that

`\tan^2A-\sin^2A=\sin^2A\tan^2A`

and assuming
`\sin A>0` and
`\tan A>0`, we end up with

`4√(\tan^2A-\sin^2A)=4\tan A\sin A`

so that

`m^2-n^2=4√(mn)`

as required.