RS Aggarwal Class 10 Math Eighth Chapter Trigonometric Identities Exercise 8A Solution
RS Aggarwal Class 10 Math New Edition Chapter 8’s First Exercise 8A Question wise Solution.
(17) (i) L.H.S = Sin6θ + Cos6θ
=(Sin6θ) + (Cos6θ)
= (Sin2θ)3 + (Cos2θ)3
= (Sin2θ + Cos2θ)3 – 3Sin2 θCos2 θ(Sin2θ + Cos2θ) [∵ a3 + b3 = (a + b)3 – 3ab (a+b)]
= 13 – 3Sin2 θCos2 θ x 1 [∵ Sin2 θ + Cos2 θ = 1]
= 1 – 3Sin2 θCos2 θ = RHS
Therefore, LHS = RHS (Proved)
(ii) LHS = Sin2θ + Cos4θ
= Sin2θ + Cos2θ x Cos2θ
= Sin2θ + Cos2θ (1 – Sin2θ) [∵Cos2θ = 1 – Sin2θ]
= Sin2θ + Cos2θ – Sin2θCos2θ
= 1 – Sin2θCos2θ [∵ Sin2θ + Cos2θ = 1)
RHS
= Cos2θ + Sin4θ
= Cos2θ + Sin2θ x Sin2 θ
= Cos2θ + Sin2θ (1 – Cos2θ) [∵Sin2 θ = 1 – Cos2 θ]
= Cos2 θ + Sin2 θ – Sin2 θCos2 θ
= 1 – Sin2 θCos2 θ [∵ Sin2θ + Cos2θ = 1)
Therefore, LHS = RHS (Proved)
(iii) LHS = Cosec4 θ – Cosec2 θ
= Cosec2 θ (Cosec2 θ – 1)
= Cosec2 θ x Cot2 θ
= (1 + Cot2 θ)Cot2 θ
= Cot2 θ + Cot4 θ = RHS
Therefore, LHS = RHS (Proved)
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