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

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

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Approved

[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