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Ex 8.3

Ex 8.3, 1

Ex 8.3, 2 Important

Ex 8.3, 3 (i) [MCQ]

Ex 8.3, 3 (ii) [MCQ] Important

Ex 8.3, 3 (iii) [MCQ] Important

Ex 8.3, 3 (iv) [MCQ]

Ex 8.3, 4 (i) Important

Ex 8.3, 4 (ii)

Ex 8.3, 4 (iii) Important

Ex 8.3, 4 (iv) Important

Ex 8.3, 4 (v) Important

Ex 8.3, 4 (vi)

Ex 8.3, 4 (vii) Important

Ex 8.3, 4 (viii)

Ex 8.3, 4 (ix) Important You are here

Ex 8.3, 4 (x)

Question 1 (i) Important Deleted for CBSE Board 2024 Exams

Question 1 (ii) Deleted for CBSE Board 2024 Exams

Last updated at Aug. 16, 2023 by Teachoo

Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) (cosec A β sin A)(sec A β cos A) = 1/(π‘ππ π΄ +cotβ‘ π΄) [Hint : Simplify LHS and RHS separately] Solving L.H.S (cosec A β sin A) (sec A β cos A) = (1/sinβ‘γ π΄γ β sinβ‘π΄ )(1/cosβ‘γ π΄γ β cosβ‘ π΄) = ((π β ππππ π¨))/sinβ‘γ π΄γ Γ ((π β ππππ π¨))/cosβ‘γ π΄γ We know that cos2 ΞΈ + sin2 ΞΈ = 1 So, cos2 ΞΈ = 1 β sin2 ΞΈ sin2 ΞΈ = 1 β cos2 ΞΈ = πππππ¨/sinβ‘γ π΄γ Γ (ππππ π¨)/cosβ‘γ π΄γ = sin A cos A Solving R.H.S 1/(π‘ππ π΄ + cotβ‘ π΄) = 1/(sinβ‘π΄/cosβ‘π΄ + cosβ‘π΄/sinβ‘π΄ ) = 1/(sinβ‘γπ΄ (sinβ‘γπ΄) + cosβ‘γπ΄ (cosβ‘γπ΄)γ γ γ γ/cosβ‘γπ΄ sinβ‘π΄ γ ) = 1/((π ππ2 π΄ + πππ 2 π΄)/cosβ‘γπ΄ sinβ‘π΄ γ ) = 1/((π ππ2 π΄ + πππ 2 π΄)/cosβ‘γπ΄ sinβ‘π΄ γ ) = sinβ‘γ π΄ . cosβ‘ π΄γ/(π ππ2 π΄ + πππ 2 π΄) As sin2 A + cos2 A = 1 = sinβ‘γ π΄ . γ cosγβ‘ π΄γ/1 = sin A cos A = L.H.S Hence proved