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aritra nandy Grade: 10
        

1+tanA=√2 sin(45+A)/cosA=√2 cos(45-A)/cosA how can 1+tanA be written as above.???


show the derivations of both 1+tanA=√2 sin(45+A)/cosA and 1+tan A =√2 cos(45-A)/cosA


need it fast.and derivation is really important.

6 years ago

Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


Its derivation is very simple...


1+tanA=1+sinA/cosA=sinA+cosA/cosA


multiply and divide by root 2


 we get


√2 sin(45+A)/cosA=√2 cos(45-A)/cosA


 




























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6 years ago
vikas askiitian expert
510 Points
										

1+ tanA = (sinA + cosA)/cosA                           (multiply divide  by 21/2)


 


1+ tanA = 21/2[sinA(1/21/2) + cosA(1/21/2) ]/cosA                  .................1


   sinpi/4 = cospi/4  =   1/21/2


now , eq  1 can be rewritten as


1+tanA = 21/2 [sinAsinpi/4+cosAcospi/4] /cosA     or  21/2[sinAcospi/4+cosAsinpi/4]/cosA


 


now we have , sinacosb+cosasinb=sina+b    &      cosacosb+sinasinb=cosa-b


using these


1+tanA = 21/2[cos(pi/4-A)]/cosA = 21/2[sin(pi/4+A)]/cosA


(pi/4 =45)


hence proved

6 years ago
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