Question

What is the value of $\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$ when $\tan \theta =\frac{4}{3}$.

• A. $\frac{16}{9}$
• B. $\frac{9}{16}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{9}{16}$

Given $\tan \theta =\frac{4}{3}$
Let's find the third side of the triangle.


$\begin{array}{cc}{\text{Hypotenuse}}^2& =3^2+4^2\\ & =9+16\\ & =25\\ \text{Hypotenuse}& =\sqrt{25}\\ & =5\end{array}$

So,
$\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}$

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

$\begin{array}{cc}\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}& =\frac{(1+\frac{4}{5})(1-\frac{4}{5})}{(1+\frac{3}{5})(1-\frac{3}{5})}\\ & =\frac{\frac{9}{5}\cdot \frac{1}{5}}{\frac{8}{5}\cdot \frac{2}{5}}\\ & =\frac{\frac{9}{25}}{\frac{16}{25}}\\ & =\frac{9}{16}\end{array}$