Question 17 (Or 2nd) - Prove that sin (1 + tan) + cos (1 + cot) = sec

Question 17 (OR 2nd question) Prove that sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ Solving LHS sin θ (1 + tan θ) + cos θ (1 + cot θ) = sin θ (1 + sin⁡𝜃/cos⁡𝜃 ) + cos θ (1 + cos⁡𝜃/sin⁡𝜃 ) = sin θ ((cos⁡𝜃 + sin⁡𝜃)/cos⁡𝜃 ) + cos θ ((sin⁡𝜃 + cos⁡𝜃)/sin⁡𝜃 ) = sin θ ((cos⁡𝜃 + sin⁡𝜃)/cos⁡𝜃 ) + cos θ ((sin⁡𝜃 + cos⁡𝜃)/sin⁡𝜃 ) Taking cos⁡𝜃 + sin⁡𝜃 common = (cos⁡𝜃 + sin⁡𝜃) (sin⁡𝜃/cos⁡𝜃 +cos⁡𝜃/sin⁡𝜃 ) = (cos⁡𝜃 + sin⁡𝜃) ((sin⁡𝜃 × sin⁡𝜃 + cos⁡𝜃 × cos⁡𝜃)/(cos⁡𝜃 sin⁡𝜃 )) = (cos⁡𝜃 + sin⁡𝜃) ((sin^2⁡𝜃 + cos^2⁡𝜃 )/(cos⁡𝜃 sin⁡𝜃 )) Since sin^2⁡𝜃 + cos^2⁡𝜃 = 1 = (cos⁡𝜃 + sin⁡𝜃) ((1 )/(cos⁡𝜃 sin⁡𝜃 )) = (cos⁡𝜃 )/(cos⁡𝜃 sin⁡𝜃 ) + (sin⁡𝜃 )/(cos⁡𝜃 sin⁡𝜃 ) = (1 )/sin⁡𝜃 + (1 )/cos⁡𝜃 = cosec θ + sec θ = RHS Since LHS = RHS Hence proved

Question 17 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards