Answer:
The expression [tex](\tan x+\cot x)^2[/tex] is same as [tex]\sec ^2x+\csc^2x[/tex]
Step-by-step explanation:
Given expression [tex](\tan x+\cot x)^2[/tex]
We have to find an equivalent fraction to the given expression.
Consider the given expression [tex](\tan x+\cot x)^2[/tex]
Using algebraic identity,
[tex](a+b)^2=a^2+b^2+2ab[/tex]
[tex](\tan x+\cot x)^2=\tan^2x+\cot^2x+2\tan x \cdot \cot x[/tex]
Using trigonometric identities,
[tex]\tan^2x=\sec ^2x-1\\\\\ \cot^2x=\csc^2x-1[/tex]
We have,
[tex]\Rightarrow \sec ^2x-1+\csc^2x-1+2\tan x \cdot \cot x[/tex]
Also, [tex]\tan x=\frac{1}{\cot x}[/tex], we have,
[tex]\Rightarrow \sec ^2x-1+\csc^2x-1+2\cdot \frac{1}{\cot x}\cdot \cot x[/tex]
On simplifying , we get,
[tex]\Rightarrow \sec ^2x-1+\csc^2x-1+2[/tex]
[tex]\Rightarrow \sec ^2x+\csc^2x[/tex]
Thus, the expression [tex](\tan x+\cot x)^2[/tex] is same as [tex]\sec ^2x+\csc^2x[/tex]
