Question #f59b1

2 Answers
Altrigeos
May 28, 2017

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

Ratnaker Mehta
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!

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