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CBSE Class 10 Answered

Question related to trigonometry
Asked by | 08 Oct, 2009, 11:55: PM
answered-by-expert Expert Answer

(1+tan2θ / 1+cot2θ ) + (1 +cot2θ/ 1+tan2θ ) =

(sec2θ / cosec2θ ) + cosec2θ/ sec2θ =

(sec4θ + cosec4θ )/(cosec2θsec2θ) =

(sin4θ + cos4θ )/(cos2θsin2θ) .......................(1)

Now, (sin2θ + cos2θ )2 = sin4θ + cos4θ + 2cos2θsin2θ

Hence, sin4θ + cos4θ = (sin2θ + cos2θ )2- 2cos2θsin2θ = 1 - 2cos2θsin2θ

Substituting this in (1),

(1 - 2cos2θsin2θ)/(cos2θsin2θ) =

sec2θ cosec2θ - 2

Regards,

Team,

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Answered by | 09 Oct, 2009, 06:49: PM
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