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.

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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Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

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

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

   sinpi/4 = cospi/4  =   1/2^1/2

now , eq  1 can be rewritten as

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

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

using these

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

(pi/4 =45)

hence proved