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the value of [2sinθ.tanθ(1-tanθ) + 2sinθ.sec 2 θ] / (1 + tanθ) 2 is a) 2cosθ /(1+ tanθ) b) 2tanθ c) 2sinθ d) 2sinθ /(1 + tanθ)

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)

Grade:11

2 Answers

AKASH GOYAL AskiitiansExpert-IITD
420 Points
13 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

AskiitiansExpert-IITD

 

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vikas askiitian expert
509 Points
13 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

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