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

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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

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

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