Question 8 & 9, Review Exercise 10

Solutions of Question 8 & 9 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove the identity sin(π4θ)sin(π4+θ)=12cos2θ.

We know that 2sinαsinβ=cos(αβ)cos(α+β) L.H.S.=sin(π4θ)sin(π4+θ)=12[2sin(π4+θ)sin(π4θ)]=12[cos((π4+θ)(π4θ))cos((π4+θ)+(π4+θ))]=12(cos2θcosπ2)=12(cos2θ0)=cos2θ2=R.H.S.

Prove that sin2(π+θ)tan(3π2+θ)cot2(3π2θ)cos2(πθ)cosec(2πθ)=cosθ.

L.H.S.=sin2(π+θ)tan(3π2+θ)cot2(3π2θ)cos2(πθ)cosec(2πθ)=(sin(2π2+θ))2tan(3π2+θ)(cot(3π2θ))2(cos(2π2θ))2cosec(2πθ)=(sinθ)2(cotθ)(tanθ)2(cosθ)2cosec(θ)=sin2θ(cosθsinθ)(sin2θcos2θ)(cos2θ)(1sinθ)=cosθ=R.H.S.

Prove that cos(90+θ)sec(θ)tan(180θ)sec(360θ)sin(180+θ)cot(90θ)=1.

L.H.S.=cos(90+θ)sec(θ)tan(180θ)sec(360θ)sin(180+θ)cot(90θ)=cos(90+θ)secθtan(290θ)sec(490θ)sin(290+θ)cot(90θ)=sinθsecθ(tanθ)secθ(sinθ)tanθ=1=R.H.S.