# the value of         [2sinθ.tanθ(1-tanθ) + 2sinθ.sec2θ] / (1 + tanθ)2     isa) 2cosθ /(1+ tanθ)b) 2tanθc) 2sinθd) 2sinθ /(1 + tanθ)

420 Points
12 years ago

Dear Student

I am taking theta=A

in the numerator put sec2A=1+tan2A

Numerator=2sinAtanA-2sinAtan2A+2sinA+2sinAtan2A

= 2sinA(1+tanA)

hence the expression will become

2sinA(1+tanA)/(1+tanA)2

=2sinA/(1+tanA)

option d

All the best.

AKASH GOYAL

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

509 Points
12 years ago

[2sin@.tan@(1-tan@) + 2sin@sec2@] / (1+tan@)2

= 2sin@ [ tan@(1-tan@) + sec2@] / (1+tan@)2

=2sin@ [ tan@ + sec2@ - tan2@ ] / (1+tan2@)2                   (sec2@ - tan2@ =1)

=2sin@ [ tan@ + 1 ] /(1+tan@)2

=2sin@ / (1+tan@)                                                       (tan@ = sin@/cos@)

option d is correct