Question

${\left[\frac{1+tan\theta }{1+cot\theta }\right]}^4$ is equal to

• A. ${\left(\frac{1+ta{n}^2\theta }{1+co{t}^2\theta }\right)}^2$
• B. ${\left(\frac{1+cot\theta }{1+tan\theta }\right)}^2$
• C. ${\left(\frac{1+tan\theta }{1-cot\theta }\right)}^2$
• D. ${\left(\frac{1-co{t}^2\theta }{1+ta{n}^2\theta }\right)}^2$

Answer 1

The correct option is A ${\left(\frac{1+ta{n}^2\theta }{1+co{t}^2\theta }\right)}^2$

${\left[\frac{1+tan\theta }{1+cot\theta }\right]}^4$

$=\left(\frac{tan\theta (1+tan\theta )}{1+tan\theta }\right){}^4$ ...($1/cot\theta =tan\theta$)

$=\left(tan\theta \right){}^4$

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

$=\left(\frac{ta{n}^2\theta (1+ta{n}^2\theta )}{(1+co{t}^2\theta )}\right)$

$=\left(\frac{(1+ta{n}^2\theta )}{(1+co{t}^2\theta )}\right){}^2$

Hence the answer is A