(c) \(\sqrt{\frac{a^2+b^2}{b}}\)
Given, \(\frac{sin\,\theta}{cos\,\theta}=\frac{a}{b}\) ⇒ tan θ = \(\frac{a}{b}\)
Also, sec2 θ = 1 + tan2θ ⇒ 2 sec θ = \(\sqrt{1+tan^2\,\theta}\)
⇒ sec θ = \(\sqrt{1+\frac{a^2}{b^2}}=\) \(\sqrt{\frac{b^2+a^2}{b^2}}=\)\(\sqrt{\frac{a^2+b^2}{b}}\)