How do you differentiate $ y = {\cot ^2}\sin \theta $ ?

Hint: Solve this equation using chain rule and should know the value for $ \sin \theta $ and $ \cot \theta $ . There is nothing that needs to be evaluated and it just involves mere substitutions of some formulas. If you know the basic formula of trigonometry, you are able to solve this problem.

Complete step-by-step answer:
To deal with the function of function problems, we have to deal with chain rule. You should know how to separate the function, so that you are able to differentiate the function properly. The formula for chain rule is,
 $ \dfrac{d}{{d\theta}}[f(g(\theta))] = \dfrac{{df(g(\theta))}}{{d\theta}}\dfrac{{dg(\theta)}}{{d\theta}} $
Take $ f(g(\theta)) = {\cot ^2}(\sin \theta ) $ and $ g(\theta) = \sin \theta $ . When we apply it to our problem we get,
  $ \dfrac{d}{{d\theta}}[{\cot ^2}(\sin \theta )] = \dfrac{{d{{\cot }^2}(\sin \theta )}}{{d\theta}}\dfrac{{d(\sin \theta )}}{{d\theta}} $
We know that, $ \dfrac{d}{{d\theta}}({\cot ^2}\sin \theta ) = - \cos e{c^2}\sin \theta $ and $ \dfrac{d}{{d\theta}}(\sin \theta ) = \cos \theta $ , these are the values we got by differentiation the given function. We separated the function using chain rule and substituting these values in the above equation, we get,
 $ \dfrac{d}{{d\theta}}[{\cot ^2}(\sin \theta )] $
 $
   = 2\cot (\sin \theta )( - \cos e{c^2}\sin \theta )(\cos \theta ) \\
   = - 2\cos \theta \cos e{c^2}(\sin \theta )\cot (\sin \theta ) \;
    $
This is our required solution.
So, the correct answer is “ $ - 2\cos \theta \cos e{c^2}(\sin \theta )\cot (\sin \theta ) $ ”.

Note: This problem is a somewhat tricky question, considering the given problem $ y = {\cot ^2}\sin \theta $ , first we need to differentiate $ {\cot ^2} $ , we get $ 2\cot $ . Then we need to consider $ \cot $ , which will be equal to $ - \cos e{c^2} $ . And at last, we need to consider $ \sin \theta $ , which is equal to $ \cos \theta $ . If you leave any one of this part, we are not going to be getting the correct solution.