Ex 8.3, 4 (vii) - Chapter 8 Class 10 Introduction to Trignometry
Last updated at April 16, 2024 by Teachoo
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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 You are here
Ex 8.3, 4 (viii)
Ex 8.3, 4 (ix) Important
Ex 8.3, 4 (x)
Question 1 (i) Important Deleted for CBSE Board 2024 Exams
Question 1 (ii) Deleted for CBSE Board 2024 Exams
Examples →
Chapter 8 Class 10 Introduction to Trignometry
Serial order wise
- Ex 8.1
- Ex 8.2
-
Ex 8.3
- Examples
- Case Based Questions (MCQ)
- NCERT Exemplar - MCQ
- sin 90 - θ, cos 90 - θ formula
Transcript
Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) (sin θ − 2 sin3 θ)/(2 cos3 θ − cos θ)=tan θ Solving L.H.S (sin θ − 2 sin3θ)/(2 cos3 θ − cosθ) = (𝒔𝒊𝒏𝜽 (1 − 2 sin2 θ))/(𝐜𝐨𝐬 𝛉(2cos2 θ − 1)) = sinθ/(cos θ) × ( (1 − 2sin2θ))/((2cos2θ − 1)) = tan θ × ( (1 − 2sin2θ))/((2𝐜𝐨𝐬𝟐𝛉 − 1)) We know that cos2 θ + sin2 θ = 1 cos2 θ = 1 – sin2 θ = tan θ × ((1 − 2sin2θ))/((2(𝟏 − 𝐬𝐢𝐧𝟐𝜽) −1) ) = tan θ × ((1 − 2sin2θ))/((2 − 2sin2θ − 1)) = tan θ × ((𝟏 − 𝟐𝐬𝐢𝐧𝟐𝛉))/((𝟏 − 𝟐𝐬𝐢𝐧𝟐𝛉) ) = tan θ × 1 = tan θ = R.H.S Hence proved
