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If sinθ = \(\frac{a}b\), find secθ + tanθ in terms of a and b.

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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 + a

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

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

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