Question

if tan teeta+ sin teeta=m and tan teeta - sin teeta = n show that ( m² - n²)² = 16mn

Answer 1

I hope it will help you

Take 16mn = 4√mn

tan(θ) + sin(θ) = m

tan(θ) - sin(θ) = n


m.n = [tan(θ) + sin(θ)].[tan(θ) - sin(θ)]

m.n = tan²(θ) - sin²(θ)

m.n = [sin²(θ)/cos²(θ)] - sin²(θ)

m.n = [sin²(θ) - sin²(θ).cos²(θ)]/cos²(θ)

m.n = sin²(θ).[1 - cos²(θ)]/cos²(θ) → recall: cos² + sin² = 1 → sin² = 1 - cos²

m.n = sin²(θ).[sin²(θ)]/cos²(θ)

m.n = sin²(θ).[sin²(θ)/cos²(θ)]

m.n = sin²(θ).tan²(θ)

√(m.n) = sin(θ).tan(θ) ← memorize this result as (i)



m² - n² = [tan(θ) + sin(θ)]² - [tanθ) - sin(θ)]²

m² - n² = [tan²(θ) + 2.tan(θ).sin(θ) + sin²(θ)] - [tan²(θ) - 2.tan(θ).sin(θ) + sin²(θ)]

m² - n² = tan²(θ) + 2.tan(θ).sin(θ) + sin²(θ) - tan²(θ) + 2.tan(θ).sin(θ) - sin²(θ)

m² - n² = 4.tan(θ).sin(θ) → recall (i)

m² - n² = 4.√(m.n)

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Answer 2

HEYA!!!!

YOUR SOLUTION IS IN THE ATTACHMENT.

#KEVIN
HOPE IT HELPS YOU ☺