Answer:
E.) 1
Step-by-step explanation:
Firstly we will solve for L.H.S.
L.H.S. =[tex]Csc^2\theta[/tex]
Since we know that [tex]Csc^2\theta[/tex] is the inverse of [tex]Sin^2\theta[/tex].
So we can say that;
[tex]csc^2\theta=\frac{1}{sin^2\theta}[/tex]
Now For R.H.S.
[tex]Cot^2\theta+1[/tex]
Since we can rewrite [tex]cot^2\theta[/tex] as [tex]\frac{cos^2\theta}{sin^2\theta}[/tex].
Now we can say that the R.H.S. is;
[tex]\frac{cos^2\theta}{sin^2\theta}+1[/tex]
Now we add the fraction and get;
[tex]\frac{cos^2\theta+sin^2\theta}{sin^2\theta}[/tex]
Now according to trigonometric identity;
[tex]cos^2\theta+sin^2\theta=1[/tex]
So, [tex]\frac{cos^2\theta+sin^2\theta}{sin^2\theta}=\frac{1}{sin^2\theta}[/tex]
Here,
[tex]csc^2\theta=\frac{1}{sin^2\theta}[/tex] and [tex]Cot^2\theta+1[/tex] = [tex]\frac{1}{sin^2\theta}[/tex]
L.H.S. = R.H.S.
Hence [tex]csc^2\theta=cot^2\theta+1[/tex]
