Question #f59b1

2 Answers
May 28, 2017

See below.

Explanation:

Given:
p=sintheta+costhetap=sinθ+cosθ
q=sectheta+cscthetaq=secθ+cscθ

Proving:
q(p^2-1)=2pq(p21)=2p
q((sintheta+costheta)^2-1)=RHSq((sinθ+cosθ)21)=RHS
q(sin^2theta+cos^2theta-1+2sinthetacostheta)=RHSq(sin2θ+cos2θ1+2sinθcosθ)=RHS
(sectheta+csctheta)(1-1+2sinthetacostheta)=RHS(secθ+cscθ)(11+2sinθcosθ)=RHS

Also:
csctheta=1/sinthetacscθ=1sinθ
sectheta=1/costhetasecθ=1cosθ

Then:
(1/costheta+1/sintheta)(2sinthetacostheta)=RHS(1cosθ+1sinθ)(2sinθcosθ)=RHS

Distribute (2sinthetacostheta)(2sinθcosθ):

(2sinthetacostheta)/costheta+(2sinthetacostheta)/sintheta=RHS2sinθcosθcosθ+2sinθcosθsinθ=RHS

2sintheta+2costheta=RHS2sinθ+2cosθ=RHS
2(sintheta+costheta)=RHS2(sinθ+cosθ)=RHS
2p=RHS2p=RHS

QED

May 28, 2017

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!