Question

Is LHS=RHS?
$\frac{1+{\cot }^2\theta }{1+{\tan }^2\theta }={\left(\frac{1+\cot \theta }{1+\tan \theta }\right)}^2$

Say true or false

• A. Yes
• B. No
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A Yes

$LHS$

$\frac{1+co{t}^2\theta }{1+ta{n}^2\theta }$
$=\frac{ta{n}^2\theta +1}{ta{n}^2\theta +1}.\frac{1}{ta{n}^2\theta }$
$=\frac{1}{ta{n}^2\theta }$
$RHS$
$=\frac{1+co{t}^2\theta +2cot\theta }{1+ta{n}^2\theta +2tan\theta }$
$=\frac{ta{n}^2\theta +1+2tan\theta }{1+ta{n}^2\theta +2tan\theta }.\frac{1}{ta{n}^2\theta }$
$=\frac{1}{ta{n}^2\theta }$
Hence
$\therefore LHS=RHS.$
Hence proved.