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