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