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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.
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
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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