Question

If $\tan \theta =\frac{3}{2},$ then find the value of $\frac{(2+2\sec \theta )(1-\sec \theta )}{(2+2cosec\theta )(1-cosec\theta )}$

• A. 1/16
• B. 27/16
• C. 1
• D. 81/16

Answer 1

The correct option is D 81/16

Given, $\tan \theta =\frac{3}{2}$

$\Rightarrow \cot \theta =\frac{2}{3}$

Now,

$\frac{(2+2\sec \theta )(1-\sec \theta )}{(2+2cosec\theta )(1-cosec\theta )}$

$=\frac{(1+\sec \theta )(1-\sec \theta )}{(1+cosec\theta )(1-cosec\theta )}$

$=\frac{1-{\sec }^2\theta }{1-cose{c}^2\theta }$

$=\frac{{\sec }^2\theta -1}{cose{c}^2\theta -1}$

$=\frac{{\tan }^2\theta }{{\cot }^2\theta }$

$=\frac{\frac{9}{4}}{\frac{4}{9}}$

$=\frac{81}{16}$