Question

If $a=\sec x-\tan x\mathrm{and}b=\mathrm{cosec}x+\cot x$, then shown that $ab+a-b+1=0$.

Answer 1

$$\begin{gathered} a=\sec x-\tan x\mathrm{And},b=\mathrm{cosec}x+\cot x \\ =\frac{1-\sin x}{\cos x}\mathrm{And},b=\frac{1+\cos x}{\sin x} \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ ab+a-b+1 \\ \left(\frac{1-\sin x}{\cos x}\right)\left(\frac{1+\cos x}{\sin x}\right)+\frac{1-\sin x}{\cos x}-\left(\frac{1+\cos x}{\sin x}\right)+1 \\ =\frac{1-\sin x+\cos x-\sin x\cos x+\sin x-{\sin }^{2}x-\cos x-{\cos }^{2}x+\sin x\cos x}{\sin x\cos x} \\ =\frac{1-{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x} \\ =0 \end{gathered}$$

Hence proved.