Question

Is LHS=RHS?
$(1+\frac{1}{{\tan }^2\theta })(1+\frac{1}{{\cot }^2\theta })=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$
Say yes or no.

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

Answer 1

Answer: (A)

Yes

$LHS=(1+\frac{1}{{\tan }^2\theta })(1+\frac{1}{{\cot }^2\theta })$
$=\left(\frac{{\tan }^2\theta +1}{{\tan }^2\theta }\right)\left(\frac{{\cot }^2\theta +1}{{\cot }^2\theta }\right)=\frac{\left({\sec }^2\theta \right)\left(cose{c}^2\theta \right)}{(\tan \theta \cot \theta {)}^2}$
$=\frac{{\sec }^2\theta cose{c}^2\theta }{1}=\frac{1}{{\cos }^2\theta }\times \frac{1}{{\sin }^2\theta }=\frac{1}{(1-{\sin }^2\theta ){\sin }^2\theta }=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }=RHS$