Question 20

Question 20 Air. 20 tan 5' tan tan 77 -0) that qnoor - -A) tan A Answers L fii)0 (iii)0 (iv)0 (v)0 (b) 0 12

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cosθ sinθ - \frac{sinθ \cos (90 ° - θ) cosθ}{\sec (90 ° - θ)} - \frac{cosθ \sin (90 ° - θ) sinθ}{cosec (90 ° - θ)} = cosθ sinθ - \frac{sinθ sinθ cosθ}{cosecθ} - \frac{cosθ cosθ sinθ}{secθ} = cosθ sinθ - \sin^{3} θ cosθ - \cos^{3} θ sinθ = cosθ sinθ - sinθ cosθ (\sin^{2} θ + \cos^{2} θ) = cosθsinθ - cosθ sinθ = 0 Note : 1) \cos (90 ° - x) = sinx 2) \sec (90 ° - x) = cosecx 3) \sin (90 ° - x) = cosx 4) cosec (90 ° - x) = secx 5) \sin^{2} x + \cos^{2} x = 1 6) \frac{1}{secx} = cosx 7) \frac{1}{cosecx} = sinx

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