Question

Question #f59b1

Answer 1

See below.

Explanation:

Given:
`p=sintheta+costheta`
`q=sectheta+csctheta`

Proving:
`q(p^2-1)=2p`
`q((sintheta+costheta)^2-1)=RHS`
`q(sin^2theta+cos^2theta-1+2sinthetacostheta)=RHS`
`(sectheta+csctheta)(1-1+2sinthetacostheta)=RHS`

Also:
`csctheta=1/sintheta`
`sectheta=1/costheta`

Then:
`(1/costheta+1/sintheta)(2sinthetacostheta)=RHS`

Distribute `(2sinthetacostheta)`:

`(2sinthetacostheta)/costheta+(2sinthetacostheta)/sintheta=RHS`

`2sintheta+2costheta=RHS`
`2(sintheta+costheta)=RHS`
`2p=RHS`

QED

Answer 2

See the Proof given in the Explanation.

Explanation:

`sintheta+costheta=p,...............(1).................."[Given]."`

`rArr (sintheta+costheta)^2=p^2.`

`rArr sin^2theta+2sinthetacostheta+cos^2theta=p^2.`

#because, sin^2theta+cos^2theta=1, :., 1+2sinthetacostheta=p^2..................(2).#

But, `sectheta+csctheta=q, ............................["Given]."`

`:. 1/costheta+1/sintheta=q.`

`:. (sintheta+costheta)/(sinthetacostheta)=q.`

`:. p/q=sinthetacostheta,............................[because, (1)].`

Subst.ing this in `(2), 1+2(p/q)=p^2, or, (q+2p)/q=p^2`

`rArr q+2p=qp^2 rArr 2p=q(p^2-1),`as desired!