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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.

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.

Grade:10

2 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

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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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

vikas askiitian expert
509 Points
10 years ago

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

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