If sin θ : cos θ :: a : b, then the value of sec θ is(a) \(\frac{\

1.	If sin θ : cos θ :: a : b, then the value of sec θ is(a) \(\frac{\sqrt{a^2+b^2}}{a}\)(b) \(\frac{b}{\sqrt{a^2+b^2}}\)(c) \(\frac{\sqrt{a^2+b^2}}{b}\)(d) \(\frac{a}{\sqrt{a^2+b^2}}\)
Answer» (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 = \(\sqrt{1+\frac{a^2}{b^2}}=\) \(\sqrt{\frac{b^2+a^2}{b^2}}=\)\(\sqrt{\frac{a^2+b^2}{b}}\)

If sin θ : cos θ :: a : b, then the value of sec θ is(a) \(\frac{\sqrt{a^2+b^2}}{a}\)(b) \(\frac{b}{\sqrt{a^2+b^2}}\)(c) \(\frac{\sqrt{a^2+b^2}}{b}\)(d) \(\frac{a}{\sqrt{a^2+b^2}}\)

Answer»

(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 = \(\sqrt{1+\frac{a^2}{b^2}}=\) \(\sqrt{\frac{b^2+a^2}{b^2}}=\)\(\sqrt{\frac{a^2+b^2}{b}}\)