Question

Simplify $\frac{3-4{\sin }^2\Theta }{{\cos }^2\Theta }$

• A. $3-{\cot }^2\Theta$
• B. $3+{\cot }^2\Theta$
• C. $3-{\tan }^2\Theta$
• D. $3+{\tan }^2\Theta$

Answer 1

The correct option is C $3-{\tan }^2\Theta$

$\frac{3-4si{n}^2\theta }{co{s}^2\theta }=\frac{3}{co{s}^2\theta }-4\frac{si{n}^2\theta }{co{s}^2\theta }$

As we know that $\frac{1}{co{s}^2\theta }=se{c}^2\theta$ and $\frac{si{n}^2\theta }{co{s}^2\theta }=ta{n}^2\theta$

$\Rightarrow 3se{c}^2\theta -4ta{n}^2\theta =3(1+ta{n}^2\theta )-4ta{n}^2\theta =3-ta{n}^2\theta$

since $se{c}^2\theta -ta{n}^2\theta =1$