Given: sinθ = \(\frac{a}b\)
To find: secθ + tanθ in terms of a and b.
Solution:
sinθ = \(\frac{a}b\) 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 = bAC2 = AB2 + BC2
b2 = AB2 + a2
AB2 = b2 - a2
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 = \(\frac{b}{\sqrt{b^2 - a^2}}\) + \(\frac{a}{\sqrt{b^2 - a^2}}\)
= \(\sqrt{\frac{b+a}{b-a}}\)