Last updated at Feb. 14, 2025 by Teachoo
Ex 8.3, 4 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. "cos A β sin A + 1" /"cos A + sin A β 1" = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A. Solving L.H.S (cosβ‘π΄ β sinβ‘π΄ + 1)/(cosβ‘π΄ + sinβ‘π΄ β 1) Since we need to use cosec and cot identity Dividing both numerator and denominator by sin A = (π/πππβ‘γ π¨γ (cosβ‘γ π΄ β sinβ‘γπ΄ + 1γ γ ))/(π/πππβ‘γ π¨γ (cosβ‘γ π΄ + sinβ‘γ π΄ β 1γ γ ) ) = (cosβ‘γ π΄γ/sinβ‘γ π΄γ β sinβ‘γ π΄γ/sinβ‘γ π΄γ + 1/sinβ‘γ π΄γ )/(cosβ‘γ π΄γ/sinβ‘γ π΄γ + sinβ‘γ π΄γ/sinβ‘γ π΄γ β 1/sinβ‘γ π΄γ ) = cotβ‘γ π΄ β 1 + πππ ππ π΄γ/cotβ‘γ π΄ + 1 β πππ ππ π΄γ = ((cotβ‘γ π΄ + πππ ππ π΄) β πγ)/((cotβ‘γ π΄ + 1 β πππ ππ π΄) γ ) = ((coπ‘β‘γ π΄ + πππ ππ π΄) β (ππππππ π¨ β ππππ π¨)γ)/((cotβ‘ π΄ + 1 β πππ ππ π΄)) = ((coπ‘β‘γ π΄ + πππ ππ π΄) β (ππ¨πππβ‘π¨ β πππ π¨)(ππ¨πππβ‘π¨ + πππ π¨)γ)/((cotβ‘ π΄ + 1 β πππ ππ π΄)) = ((coπ‘β‘γ π΄ + πππ ππ π΄) [π β (πππππ π¨ β πππ π¨ )]γ)/([cotβ‘ π΄ + 1 β πππ ππ π΄]) = ((coπ‘β‘γ π΄ + πππ ππ π΄) [1 β πππ ππ π΄ + πππ‘ π΄]γ)/([cotβ‘ π΄ + 1 β πππ ππ π΄]) = ((coπ‘β‘γ π΄ + πππ ππ π΄)[cotβ‘ π΄ + 1 β πππ ππ π΄]γ)/([cotβ‘ π΄ + 1 β πππ ππ π΄]) = cot A + cosec A = R.H.S Hence proved
Ex 8.3
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 You are here
Ex 8.3, 4 (vi)
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
Question 1 (ii)
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo