Suggested: sina+sin2a=1 then cos2a+cos4a - sina+sin2a/1+cosa+cos2a=tana - sina+sin2a/cosa-cos2a=cota/2 - sin2a+sin2b+sin2c=4 sina sinb sinc - sin2a+sin2b+sin2c/sina+sinb+sinc - sin2a-sin2b/sinacosa-sinbcosb=tan(a+b) - if sina=4/5 and cosb=12/13 find the value of sin2a - sin2a+sin2b+sin2c=4 sina sinb sinc in telugu - sina+cosa/sina-cosa+sina-cosa/sina+cosa=2/sin2a-cos2a - sin2a+sin5a-sina/cos2a+cos5a+cos2a=tan2a - if sina=3/5 then find sin2a - if sina-cosa=0 then the value of sin4a-sin2a is - sina+cosa/sina-cosa+sina-cosa/sina+cosa=2/sin2a-1 - sina+sin2a Browse related:
privacy contact
Copyright 2017 bapse
bapse is powered by Google and Youtube technologies
Thank you, for using bapse.
This site was designed and coded by Michael DeMichele,
and is being used as a portfolio demonstration. View more.