If \[ a=\sec x-\tan x \] and \[ b=\cosec x+\cot x \] Show that \[ ab+a-b+1=0 \]

Solution:

\[ a=\sec x-\tan x = \frac{1-\sin x}{\cos x} \]

\[ b=\cosec x+\cot x = \frac{1+\cos x}{\sin x} \]

Now,

\[ ab+a-b+1 \]

\[ = \frac{(1-\sin x)(1+\cos x)} {\sin x\cos x} + \frac{1-\sin x}{\cos x} – \frac{1+\cos x}{\sin x} +1 \]

Taking LCM \( \sin x\cos x \),

\[ = \frac{ (1-\sin x)(1+\cos x) +\sin x(1-\sin x) -\cos x(1+\cos x) +\sin x\cos x } {\sin x\cos x} \]

\[ = \frac{ 1+\cos x-\sin x-\sin x\cos x +\sin x-\sin^2 x -\cos x-\cos^2 x +\sin x\cos x } {\sin x\cos x} \]

\[ = \frac{ 1-\sin^2 x-\cos^2 x } {\sin x\cos x} \]

\[ = \frac{1-1}{\sin x\cos x} \]

\[ =0 \]

Hence proved.