If tanA+sinA=p and tanA-sinA=q. Then proove that `p^2-q^2=4sqrt (pq)` ?

Here,
`tanA+sinA=p`

`tanA-sinA=q`

`L.H.S=p^2-q^2`

`=(tanA+sinA)^2-(tanA-sinA)^2`

`=(tanA+cancelsinA+tanA-cancelsinA)(canceltanA+sinA-canceltanA+sinA)`

`=(2tanA)cdot(2sinA)`

`=4tanAsinA`

`=4sqrt(tan^2Asin^2A)`

`=4sqrt((sec^2A-1)sinA)``color(blue)([[As,sec^2theta-tan^2theta=1]])`

`=4sqrt(sec^2A.sin^2A-sin^2A)`

`=4sqrt(sin^2A/(cos^2A)-sin^2A)`

`=4sqrt(tan^2A-sin^2A)`

`=4sqrt((tanA+sinA)(tanA-sinA))`

`=4sqrt(pq)``color(green)([As.tanA+sinA=pandtanA-sinA=q])`

`=R.H.S`

`tanA+sinA=p......(1), and, tanA-sinA=q......(2)`.

`:. (1)+(2) rArr 2tanA=p+q, or, tanA=(p+q)/2`.

`"Similarly, "sinA=(p-q)/2`.

`:. cotA=2/(p+q), and cscA=2/(p-q)`.

But, `csc^2A-cot^2A=1`,

`rArr 4/(p-q)^2-4/(p+q)^2=1, or,`.

`[4{(p+q)^2-(p-q)^2}]/{(p+q)^2(p-q)^2}=1`.

`rArr (4*4pq)/(p^2-q^2)^2=1, i.e.,`

`(p^2-q^2)^2=16pq`,

`rArr p^2-q^2=4sqrt(pq),` as desired!

Enjoy Maths.!

`RHS=4sqrt(pq)`

`=4sqrt((tanA+sinA)(tanA-sinA))`

`=4sqrt(tan^2A-sin^2A)`

`=4sqrt(sin^2A/cos^2A-sin^2A)`

`=4sqrt(sin^2Axxsec^2A-sin^2A)`

`=4sqrt(sin^2A(sec^2A-1)`

`=4sqrt(sin^2Atan^2A)`

`=4tanAsinA`

`=(tanA+sinA)^2-(tanA-sinA)^2`

`=p^2-q^2=RHS`