Question:
If $a=\sec [x-\tan x$ and $b=\operatorname{cosec} x+\cot x$, then shown that $a b+a-b+1=0$.
Solution:
$a=\sec x-\tan x \quad$ And,$b=\operatorname{cosec} x+\cot x$
$=\frac{1-\sin x}{\cos x} \quad$ And, $b=\frac{1+\cos x}{\sin x}$
Now, we have :
$a b+a-b+1$
$\left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\cos x}{\sin x}\right)+\frac{1-\sin x}{\cos x}-\left(\frac{1+\cos x}{\sin x}\right)+1$
$=\frac{1-\sin x+\cos x-\sin x \cos x+\sin x-\sin ^{2} x-\cos x-\cos ^{2} x+\sin x \cos x}{\sin x \cos x}$
$=\frac{1-\sin ^{2} x-\cos ^{2} x}{\sin x \cos x}$
= 0
Hence proved.
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