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

Let us consider a right triangle ABC, right-angled at point B

$cot\ θ = \frac{BC}{AB} = \frac{7}{8}$

If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in $ΔABC$, we obtain
$AC ^2 = AB ^2 + BC^ 2$
$= (8k) ^2 + (7k)^ 2$
$= 64k^ 2 + 49k ^2$
$= 113k^ 2$
$AC =\sqrt{113} k$

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

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

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

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

$=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}$

$=\frac{\frac{49}{113}}{\frac{64}{113}}=\frac{49}{64}$

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