Answer:
It is an identity, proved below.
Step-by-step explanation:
I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.
First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).
[tex]\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}[/tex]
csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.
[tex]\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}[/tex]
Another identity is:
[tex]\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}[/tex]
Therefore:
[tex]\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}[/tex]
Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.
[tex]\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}[/tex]
Another identity:
[tex]\displaystyle \large{\sin^2x+\cos^2x=1}[/tex]
Therefore:
[tex]\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}\longrightarrow \boxed{ \frac{1}{\sin^2x}={\frac{1}{\sin^2x}}}[/tex]
Hence proved, this is proof by using identity helping to find the specific identity.
