1 +tanAtan2A=sec 2A solve it plzzzzzz
Answers
Answered by
1
=1 + tanA.tan2A
=(tan2A -tanA)/tanA
=(tan2A/tanA)-1
=[2/(1-tan*2A)]-1
=[2cos*2A/(cos*2A-sin*2A)]-1
=(2cos*2A -cos*2A-sin*2A)/(cos*2A-sin*2A)
=1/cos2A
=sec2A
hence prove
=(tan2A -tanA)/tanA
=(tan2A/tanA)-1
=[2/(1-tan*2A)]-1
=[2cos*2A/(cos*2A-sin*2A)]-1
=(2cos*2A -cos*2A-sin*2A)/(cos*2A-sin*2A)
=1/cos2A
=sec2A
hence prove
ahanasharma2:
in first step we divide it by tanA
ok
than ( tanA-tan2A)
then take common tanA common
then solve it
i hope this can help you
Similar questions