Answer
$(1-\tan x)(1-\cot x)=2-\sec x\csc x$
Work Step by Step
$(1-\tan x)(1-\cot x)=2-\sec x\csc x$
Evaluate the product on the left side:
$1-\cot x-\tan x+\tan x\cot x=2-\sec x\csc x$
Substitute $\tan x$ by $\dfrac{\sin x}{\cos x}$ and $\cot x$ by $\dfrac{\cos x}{\sin x}$:
$1-\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}+\Big(\dfrac{\sin x}{\cos x}\Big)\Big(\dfrac{\cos x}{\sin x}\Big)=2-\sec x\csc x$
Evaluate the product on the left side:
$1-\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}+\dfrac{\sin x\cos x}{\sin x\cos x}=2-\sec x\csc x$
$1+1-\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}=2-\sec x\csc x$
$2-\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}=2-\sec x\csc x$
Take out common factor $-1$ from the fractions on the left side:
$2-\Big(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\Big)=2-\sec x\csc x$
Evaluate the sum inside the parentheses:
$2-\dfrac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}=2-\sec x\csc x$
Substitute $\sin^{2}x+\cos^{2}x$ by $1$:
$2-\dfrac{1}{\sin x\cos x}=2-\sec x\csc x$
Rewrite the fraction on the left side as $\Big(\dfrac{1}{\sin x}\Big)\Big(\dfrac{1}{\cos x}\Big)$:
$2-\Big(\dfrac{1}{\sin x}\Big)\Big(\dfrac{1}{\cos x}\Big)=2-\sec x\csc x$
Since $\dfrac{1}{\sin x}=\csc x$ and $\dfrac{1}{\cos x}=\sec x$, the identity is proved:
$2-\sec x\csc x=2-\sec x\csc x$
