Question

What is the value of $\frac{(1+\tan \theta )(1+\cot \theta )}{(1-\tan \theta )(1-\cot \theta )}$ when $\sin \theta =\frac{8}{10}$.

Answer 1

Given $\sin \theta =\frac{8}{10}$
Let's find the third side of the triangle.


$\begin{array}{cc}{10}^2& =8^2+{\text{Base}}^2\\ 100& =64+{\text{Base}}^2\\ {\text{Base}}^2& =100-64\\ & =36\\ \text{Base}& =\sqrt{36}\\ & =6\end{array}$

So,
$\tan \theta =\frac{8}{6}$ and $\cot \theta =\frac{6}{8}$

Substitute the values of $\tan \theta$ and $\cot \theta )$ into $\frac{(1+\tan \theta )(1+\cot \theta )}{(1-\tan \theta )(1-\cot \theta )}$.

$\begin{array}{cc}\frac{(1+\tan \theta )(1+\cot \theta )}{(1-\tan \theta )(1-\cot \theta )}& =\frac{(1+\frac{4}{3})(1+\frac{3}{4})}{(1-\frac{4}{3})(1-\frac{3}{4})}\\ & =\frac{\frac{7}{3}\cdot \frac{7}{4}}{\frac{-1}{3}\cdot \frac{1}{4}}\\ & =\frac{\frac{49}{12}}{-\frac{1}{12}}\\ & =-49\end{array}$