# Ex 8.3, 4 (vi) - Chapter 8 Class 10 Introduction to Trignometry

Last updated at April 16, 2024 by Teachoo

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) You are here

Ex 8.3, 4 (vii) Important

Ex 8.3, 4 (viii)

Ex 8.3, 4 (ix) Important

Ex 8.3, 4 (x)

Question 1 (i) Important Deleted for CBSE Board 2025 Exams

Question 1 (ii) Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) β((1 + sinβ‘π΄ )/(1 βγ sinγβ‘π΄ )) = sec A + tan A Solving L.H.S β((π + πππβ‘π¨ )/(π βγ πππγβ‘π¨ )) Rationalizing denominator Multiplying (1 + sin A) in numerator and denominator = β(((π + π¬π’π§β‘π¨)(π + πππβ‘γπ¨)γ )/((π β π¬π’π§β‘π¨)(π + πππβ‘γπ¨)γ )) = β(((1 + sinβ‘π΄ )2 )/(12 β π ππ2π΄)) = β(((1 + sinβ‘π΄ )2 )/(1 β π ππ2π΄)) =β(((1 + sinβ‘π΄)2 )/(ππππ π¨)) =β(((1 + sinβ‘π΄ )/(πππ π΄))^2 ) = (π + πππβ‘γ π¨γ)/πππβ‘γ π¨γ = 1/cosβ‘γ π΄γ + sinβ‘γ π΄γ/cosβ‘γ π΄γ = sec A + tan A = R.H.S Hence proved