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```        the value of
[2sinθ.tanθ(1-tanθ) + 2sinθ.sec2θ] / (1 + tanθ)2     is
a) 2cosθ /(1+ tanθ)
b) 2tanθ
c) 2sinθ
d) 2sinθ /(1 + tanθ)```
6 years ago

419 Points
```										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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```
6 years ago
510 Points
```										[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
```
6 years ago
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