Question

# If $\cot \theta = \dfrac{7}{8}$, evaluate1. $\dfrac{{(1 + \sin \theta )(1 - \sin \theta )}}{{(1 + \cos \theta )(1 - \cos \theta )}}$2. ${\cot ^2}\theta$

Hint: Here we have to use the formulae of trigonometric ratio and pythagoras theorem to find the hypotenuse.

Given, $\cot \theta = \dfrac{7}{8}$
We know that by trigonometric ratio that $\cot \theta = \dfrac{b}{p}$where (b=base, p=perpendicular)
$\Rightarrow \cot \theta = \dfrac{7}{8}{\text{ = }}\dfrac{{\text{b}}}{{\text{p}}}(\therefore p = 8,b = 7)$

Now from Pythagoras theorem we find h=Hypotenuse,
$\therefore {h^2} = {p^2} + {b^2} \\ \Rightarrow {h^2} = {8^2} + {7^2} \\ \Rightarrow h = \sqrt {64 + 49} \\ \Rightarrow h = \sqrt {113} \\$
And now we know that by trigonometric ratio,
$\sin \theta = \dfrac{p}{h} = \dfrac{8}{{\sqrt {113} }} \\ \cos \theta = \dfrac{b}{h} = \dfrac{7}{{\sqrt {113} }} \\$
For (i)
$\dfrac{{(1 + \sin \theta )(1 - \sin \theta )}}{{(1 + \cos \theta )(1 - \cos \theta )}}$
We have ${a^2} - {b^2} = (a + b)(a - b)$
Similarly,
$1 - {\sin ^2}\theta = (1 + \sin \theta )(1 - \sin \theta ) \\ 1 - {\cos ^2}\theta = (1 + \cos \theta )(1 - \cos \theta ) \\$
$\Rightarrow \dfrac{{(1 + \sin \theta )(1 - \sin \theta )}}{{(1 + \cos \theta )(1 - \cos \theta )}} = \dfrac{{1 - {{\sin }^2}\theta }}{{1 - {{\cos }^2}\theta }} = \dfrac{{1 - {{\left( {\dfrac{8}{{\sqrt {113} }}} \right)}^2}}}{{1 - {{\left( {\dfrac{7}{{\sqrt {113} }}} \right)}^2}}} = \dfrac{{(113 - 64)}}{{(113 - 49)}} = \dfrac{{49}}{{64}}$
For (ii)
Given,
$\cot \theta = \dfrac{7}{8}$
We have to find ${\cot ^2}\theta$
${\cot ^2}\theta = {(\cot \theta )^2} = {\left( {\dfrac{7}{8}} \right)^2} = \dfrac{{49}}{{64}}$

Note: Whenever such type of questions are given we can solve it by two way (i) trigonometric ratio (ii) by trigonometric identities