Answer:
The verification is in the explanation.
Step-by-step explanation:
I'm going to start with the left and play with it til I get the right.
First step: Expand the binomial square using [tex](u+v)^2=u^2+2uv+v^2[/tex].
[tex](\tan(x)+\cot(x))^2[/tex]
[tex]\tan^2(x)+2\tan(x)\cot(x)+\cot^2(x)[/tex]
Second step: Realize [tex]\tan(x)[/tex] and [tex]cot(x)[/tex] are reciprocal functions. Their product will be 1.
[Side note: [tex]\tan(x)\cot(x)=\frac{\sin(x}{\cos(x)}\frac{\cos(x)}{sin(x)}=\frac{\sin(x)\cos(x)}{\sin(x)\cos(x)}=1[/tex].]
[tex]\tan^2(x)+2(1)+\cot^2(x)[/tex]
[tex]\tan^2(x)+2+\cot^2(x)[/tex]
Third step: Recall the Pythagorean Identities [tex]\tan^2(x)+1=sec^2(x)[/tex] and [tex]1+cot^2(x)=csc^2(x)[/tex]:
[tex]\tan^2(x)+(1+1)+\cot^2(x)[/tex]
[tex]\tan^2(x)+1+1+\cot^2(x)[/tex]
[tex](\tan^2(x)+1)+(1+\cot^2(x))[/tex]
[tex]\sec^2(x)+\csc^2(x)[/tex]
