Question

$\frac{2{\cot }^2\theta +5{\tan }^2\theta +7}{cose{c}^2\theta +{\tan }^2\theta +1}-2{\cos }^2\theta$ is equal to

• A. ${\sin }^2\theta$
• B. $2{\sin }^2\theta$
• C. $5{\sin }^2\theta$
• D. $4{\cos }^2\theta$

Answer 1

The correct option is C $5{\sin }^2\theta$

Consider: $\frac{2{\cot }^2\theta +5{\tan }^2\theta +7}{cose{c}^2\theta +{\tan }^2\theta +1}-2{\cos }^2\theta$

$=\frac{2(cose{c}^2\theta -1)+5{\tan }^2\theta +7}{cose{c}^2\theta +{\tan }^2\theta +1}-2{\cos }^2\theta$ $(\because 1+{\cot }^2\theta =cose{c}^2\theta )$

$=\frac{2cose{c}^2\theta -2+5{\tan }^2\theta +7}{cose{c}^2\theta +{\sec }^2\theta }-2{\cos }^2\theta$ $(\because 1+{\tan }^2\theta ={\sec }^2\theta )$

$=\frac{2cose{c}^2\theta +5(1+{\tan }^2\theta )}{cose{c}^2\theta +{\sec }^2\theta }-2{\cos }^2\theta$

$=\frac{2cose{c}^2\theta +5{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }-2{\cos }^2\theta$

$=\frac{\frac{2}{{\sin }^2\theta }+\frac{5}{{\cos }^2\theta }}{\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }}-2{\cos }^2\theta$

$=\frac{\frac{2{\cos }^2\theta +5{\sin }^2\theta }{{\sin }^2\theta {\cos }^2\theta }}{\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta {\sin }^2\theta }}-2{\cos }^2\theta$

$=2{\cos }^2\theta +5{\sin }^2\theta -2{\cos }^2\theta$ $(\because {\cos }^2\theta +{\sin }^2\theta =1)$

$=5{\sin }^2\theta$

Hence, the correct answer is option (c).