**Given: **sinθ = \(\frac{a}b\)

**To find: **secθ + tanθ in terms of a and b.

**Solution:**

**sin**θ = \(\frac{a}b\) sinθ = \(\frac{P}H\)

We will construct a right angled triangle, right angled at B such that,**∠**BAC = θ

Perpendicular=BC=a and hypotenuse = AC = bAC^{2} = AB^{2} + BC^{2}

b^{2} = AB^{2} + a^{2 }

AB^{2} = b^{2} - a^{2}

AB = \(\sqrt{b^2 - a^2}\)

^{}

Use the formula:cosθ = \(\frac{B}H\),

tanθ = \(\frac{P}H\),

secθ = \(\frac{H}B\)

**Solve,**

sinθ = \(\frac{a}b\)

⇒ cosθ = \(\frac{\sqrt{b^2 - a^2}}{b}\)

or secθ = \(\frac{b}{\sqrt{b^2 - a^2}}\)

and tanθ = \(\frac{b}{\sqrt{b^2 - a^2}}\)

secθ + tanθ = \(\frac{b}{\sqrt{b^2 - a^2}}\) + \(\frac{a}{\sqrt{b^2 - a^2}}\)

= \(\sqrt{\frac{b+a}{b-a}}\)