Question
Tue March 23, 2010 By: Vivek Suresh Kumar

Trignometry

Expert Reply
Wed March 24, 2010

(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,

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