Question

If $cosec\theta =\frac{5}{4}$, then find the value of $\frac{(1+\tan \theta )(1-\tan \theta )}{(1+\cot \theta )(1-\cot \theta )}$

Answer 1

We know that, ${\cot }^2\theta =cose{c}^2\theta -1={\left[\frac{5}{4}\right]}^2-1=\frac{25}{16}-1$

$\Rightarrow {\cot }^2\theta =\frac{25-16}{16}$

$\Rightarrow {\cot }^2\theta =\frac{9}{16}$

$\Rightarrow \cot \theta =\frac{3}{4}=\Rightarrow \tan \theta =\frac{4}{3}$

$\frac{(1+\tan \theta )(1-\tan \theta )}{(1+\cot \theta )(1-\cot \theta )}=\frac{[1+\frac{4}{3}][1-\frac{4}{3}]}{[1+\frac{3}{4}][1-\frac{3}{4}]}=\frac{\left[\frac{7}{3}\right]\left[\frac{-1}{3}\right]}{\left[\frac{7}{4}\right]\left[\frac{1}{4}\right]}=\frac{\frac{-7}{9}}{\frac{7}{16}}=\frac{-16}{9}$