trigonometry
Asked by
| 8th Oct, 2009,
11:27: PM
(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
Regards,
Team,
TopperLearning.
Answered by
| 9th Oct, 2009,
07:25: PM
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