Question

sec A+ tanA=p then tan A=​

Answer 1

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