If a= - and b=cosec x+ , then show that ab+a - b+ 1=0.

Question

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

✓

Answer

$\text{L.H.S}=\text{ab + a} - \text{b + }1$
$=(\sec\text{x}-\tan\text{x})(\text{cosec+}\cot)+\sec\text{x}-\tan\text{x}-\text{cosec }\text{x}-\cot\text{x}+1$
$=\Big(\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\Big)\Big(\frac{1}{\sin\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+1$
$=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{1}{\cos\text{x}}\times\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\\\ \ \ \times\frac{1}{\sin\text{x}}-\tan\text{x}\times\cot\text{x}+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}$
$=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{1}{\sin\text{x}}-\frac{1}{\cos\text{x}}-1+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+1$
$=\frac{1}{\sin\text{x}\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}$
$=\frac{1\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}.\cos\text{x}}$
$=\frac{1-(\cos^2\text{x}+\sin^2\text{x})}{\sin\text{x}.\cos\text{x}}$
$=\frac{1-1}{\sin\text{x}.\cos\text{x}}=0$
$=\text{R.H.S. Hence Proved}$