Tue March 23, 2010 By: Vivek Suresh Kumar

Trignometry

1 Comments

7 years ago

(b) (2-z)/z


Consider (sin2θ + cos2θ)2 = sin4θ + 2sin2θcos2θ + cos4θ


1 = sin4θ + 2sin2θcos2θ + cos4θ


1 = (1/x2) + 2/xy + (1/y2)      ... (1)


Now,


z= 1/(1-sin2θcos2θ) = 1/(1-(1/xy))


1/xy = (z-1)/z


Put this in (1),


1 = (1/x2) + 2/xy + (1/y2)


1 = (1/x2) + 2(z-1)/z + (1/y2)


(1/x2) + (1/y2) = 1 - 2(z-1)/z


= (2-z)/z.


Regards,


Team,


TopperLearning.

Kasim Razak 2 2 days ago
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Kasim Razak 3 2 days ago
Good. I like this video/chapter/test/question/greeting
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