sec A+ tanA=p then tan A=
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Step-by-step explanation:
secA+tanA = p…………………(i)

I am using above right-angled triangle to find Trigonometric Identities :
From Pythagoras Theorem, we know that,a2+b2=c2……………..(ii),a2+b2=c2……………..(ii)
Now, sin A= a/c……………………(iii)
cos A= b/c/…………………………(iv)
Adding the squares of equ (iii) and equ (iv), we get
sin2A+cos2Asin2A+cos2A
=a2/c2+b2/c2=a2/c2+b2/c2
=(a2+b2)/c2————−=(a2+b2)/c2————−[Use equ (ii)]
=c2/c2=c2/c2
=1…………………………………..(v)=1…………………………………..(v)
Now,sec2A−tan2ANow,sec2A−tan2A
=1/cos2A−sin2A/cos2A=1/cos2A−sin2A/cos2A
=(1−sin2A)/cos2A—————−=(1−sin2A)/cos2A—————−[Use equ (v)]
=cos2
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