MCQ
$\text{If }\text{A},\text{B},\text{C}\text{ are in }\text{A}.\text{P.},\text{than}\frac{\sin\text{A}-\sin\text{C}}{\cos\text{C}-\cos\text{A}}=$
- A$\tan\text{B}$
- B$\cot\text{B}$
- C$\tan2\text{B}$
- DNone of these
Solution:
Since A,B and C are in A.P,
$\text{B}-\text{A}=\text{C}-\text{B}$
$\text{Or },2\text{B}=\text{A}+\text{C}$
$\frac{\sin\text{A}-\sin\text{C}}{\cos\text{C}-\cos\text{A}}$
$=\ \frac{2\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-2\sin\Big(\frac{\text{C+A}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$ $\Big[\because\ \sin\text{A}-\sin\text{B}=2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and }\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$
$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$
$=\ \frac{\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)}$
$=\ \frac{\cos\text{B}}{\sin\text{B}}$
$=\ \cot\text{B}$
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