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.
Question 8
Prove the identity sin(π4−θ)sin(π4+θ)=12cos2θ.
Solution
We know that 2sinαsinβ=cos(α−β)−cos(α+β) L.H.S.=sin(π4−θ)sin(π4+θ)=12[2sin(π4+θ)sin(π4−θ)]=12[cos((π4+θ)−(π4−θ))−cos((π4+θ)+(π4+θ))]=12(cos2θ−cosπ2)=12(cos2θ−0)=cos2θ2=R.H.S.
Question 9(i)
Prove that sin2(π+θ)tan(3π2+θ)cot2(3π2−θ)cos2(π−θ)cosec(2π−θ)=cosθ.
Solution
L.H.S.=sin2(π+θ)tan(3π2+θ)cot2(3π2−θ)cos2(π−θ)cosec(2π−θ)=(sin(2π2+θ))2tan(3π2+θ)(cot(3π2−θ))2(cos(2π2−θ))2cosec(2π−θ)=(−sinθ)2(−cotθ)(tanθ)2(−cosθ)2cosec(−θ)=sin2θ(−cosθsinθ)(sin2θcos2θ)(cos2θ)(−1sinθ)=cosθ=R.H.S.
Question 9(ii)
Prove that cos(90∘+θ)sec(−θ)tan(180∘−θ)sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)=−1.
Solution
L.H.S.=cos(90∘+θ)sec(−θ)tan(180∘−θ)sec(360∘−θ)sin(180∘+θ)cot(90∘−θ)=cos(90∘+θ)secθtan(2⋅90∘−θ)sec(4⋅90∘−θ)sin(2⋅90∘+θ)cot(90∘−θ)=sinθ⋅secθ⋅(−tanθ)secθ(sinθ)tanθ=−1=R.H.S.
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