Question

If \(cot\  \theta = \frac{7}{8},\) evaluate:

(i) \(\frac{(1 + sin\  \theta)(1 – sin θ)}{(1+cos θ)(1-cos θ)}\)

(ii) \(cot^2\) \(θ\)

Answer 1

Consider right triangle ABC, with the right angle at B.

\(cot\ θ = \frac{BC}{AB} = \frac{7}{8}\)

Let BC = 7k. Then AB = 8k, where k is a positive integer.
Using the Pythagorean theorem in \(ΔABC\), we get:
\(AC ^2 = AB ^2 + BC^ 2\)
\(= (8k) ^2 + (7k)^ 2\)
\(= 64k^ 2 + 49k ^2\)
\(= 113k^ 2\)
\(AC =\sqrt{113} k\)

\(sin\ θ = \frac{\text{Opposite} \ \text{Side}}{\text{Hypotenuse} }=\frac{AB}{AC} = \frac{ 8k}{\sqrt{113} k} =\frac{ 8}{\sqrt{113}}\) and

\(cos\ θ =\frac{ \text{Adjacent}\ \text{ Side}}{\text{Hypotenuse }}= \frac{BC}{AC} =\frac{ 7k}{\sqrt{113} k} =\frac{ 7}{\sqrt{113}}\)

(i) \(\frac{(1 + sin θ)(1 – sin θ)}{(1+cos θ)(1-cos θ)}=\frac{(1-sin^2 θ)}{(1-cos^2 θ)}\)

\(=\frac{(1-(\frac{8}{\sqrt{113}})^2)}{(1-(\frac{7}{\sqrt{113}})^2)}\)

\(=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}\)

\(=\frac{\frac{113-64}{113}}{\frac{113-49}{113}}=\frac{\frac{49}{113}}{\frac{64}{113}}=\frac{49}{64}\)
 

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(ii) \(cot^2\ θ = (cot\ θ)^2 =(\frac{7}{8})^2=\frac{49}{64}\)