Question

Prove that tan^3/1+tan^2 + cot^3/1+cot^2=sec.cosec-2sin.cos

Answer 1

Answer :

L.H.S. = tan³θ/(1 + tan²θ) + cot³θ/(1 + cot²θ)

= tan³θ/(sec²θ) + cot³θ/(cosec²θ),
since sec²θ - tan²θ = 1 & cosec²θ - cot²θ = 1

= tan³θ cos²θ + cot³θ sin²θ

= (sin³θ/cos³θ) cos²θ + (cos³θ/sin³θ) sin²θ,
since tanθ = sinθ/cosθ & cotθ = cosθ/sinθ

= sin³θ/cosθ + cos³θ/sinθ

= (sinθ sin²θ)/cosθ + (cosθ cos²θ)/sinθ

= {sinθ (1 - cos²θ)}/cosθ + {cosθ (1 - sin²θ)}/sinθ,
since sin²θ + cos²θ = 1

= (sinθ - sinθ cos²θ)/cosθ + (cosθ - cosθ sin²θ)/sinθ

= sinθ/cosθ - sinθ cosθ + cosθ/sinθ - cosθ sinθ

= (sinθ/cosθ + cosθ/sinθ) - 2 sinθ cosθ

= (sin²θ + cos²θ)/(sinθ cosθ) - 2 sinθ cosθ

= 1/(sinθ cosθ) - 2 sinθ cosθ,
since sin²θ + cos²θ = 1

= cosecθ secθ - 2 sinθ cosθ,
since 1/sinθ = cosecθ & 1/cosθ = secθ

= secθ cosecθ - 2 sinθ cosθ

= R.H.S (Proved)

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