Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

Trignometry

Asked by viveksureshkumar 23rd March 2010, 12:46 PM
Answered by Expert
Answer:

(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.

Answered by Expert 24th March 2010, 8:37 AM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp