Question

to verify the identity, start with the more complicated side and transform it to look like the other side. choos
sin x + sin x cot²x
= sin x (□)
(do not simplify.)
= sin x (□)
= □
(do not simplify.)
= csc x

Explanation:

Step1: Factor out $\sin x$

We have the expression $\sin x+\sin x\cot^{2}x$. By the distributive property \(a + ab=a(1 + b)\), here \(a=\sin x\) and \(b = \cot^{2}x\). So, \(\sin x+\sin x\cot^{2}x=\sin x(1+\cot^{2}x)\)

Step2: Use the Pythagorean identity

We know the Pythagorean identity \(1+\cot^{2}x=\csc^{2}x\). Substituting this into the previous expression, we get \(\sin x(1 + \cot^{2}x)=\sin x\csc^{2}x\)

Step3: Use the reciprocal identity

Since \(\csc x=\frac{1}{\sin x}\), then \(\sin x\csc^{2}x=\sin x\times\frac{1}{\sin^{2}x}=\frac{1}{\sin x}\) and \(\frac{1}{\sin x}=\csc x\)

Answer:

First blank: \(1+\cot^{2}x\)；Second blank: \(\csc^{2}x\)；Third blank: \(\sin x\csc^{2}x\)