Wednesday, July 3, 2019

Q. Prove that: sec2A+tan2A=tan(45+A)


L.H.S.
= 1 / cos(2a) + sin(2a) / cos(2a)
= (1 + sin(2a)) / (cos(2a))
= (sin^2(a) + cos^2(a) + 2 * sin(a) * cos(a)) / (cos^2(a) - sin^2(a))
= (sin(a) + cos(a))^2 / ((cos(a) - sin(a)) * (cos(a) + sin(a)))
= (cos(a) + sin(a)) / (cos(a) - sin(a))
= cos(a) * (1 + sin(a) / cos(a)) / (cos(a) * (1 - sin(a)/cos(a)))
= (1 + tan(a)) / (1 - tan(a))
= (tan(45) + tan(a)) / (1 - tan(a) * tan(45))
= tan(a + 45)