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CBSE Class 10 Answered

trigonometry
Asked by | 08 Oct, 2009, 11:27: PM
answered-by-expert Expert Answer

(tan3θ / 1+tan2θ)  +( cot 3θ / 1+cot2θ ) =

(tan3θ / sec2θ)  +( cot 3θ / cosec2θ ) =

(sin3θ / cos θ)  +( cos 3θ / sin θ ) =

(sin4θ + cos 4θ)/(sin θcos θ) ..........................(1)

Now, (sin2θ + cos 2θ)2 = sin4θ + cos 4θ + 2sin2 θcos2 θ

Therefore, sin4θ + cos 4θ = (sin2θ + cos 2θ)2- 2sin2 θcos2 θ = 1 - 2sin2 θcos2 θ

Substituting this in (1),

(1 - 2sin2 θcos2 θ)/(sin θcos θ) =

sec θ cosec θ - 2sin θ cos θ

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Answered by | 09 Oct, 2009, 07:25: PM

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