Question

Sin theta = a/b then prove sec theta + tan theta = whole under √b+a/b-a

Answer 1

sinA=a/b cosA=root(b^2-a^2)/b
secA+tanA
=1/cosA + sin A/cos A
=(1+sinA )/cosA
=(1+a/b) / (root (b^2-a^2) /b)
=(b+a)/root(b+a)(b-a)
=whole root (b +a)/(b-a) =RHS