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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II1 Mark
MCQ
If $\tan A-\tan B=x$ and $\cot B-\cot A=y$, then $\cot (A-B)=$
  • A
    $\frac{1}{y}-\frac{1}{x}$
  • B
    $\frac{1}{x}-\frac{1}{y}$
  • ✓
    $\frac{1}{x}+\frac{1}{y}$
  • D
    $\frac{x y}{x-y}$
✓

Answer

Correct option: C.
$\frac{1}{x}+\frac{1}{y}$
(c) $\frac{1}{x}+\frac{1}{y}$
Hint:
$\begin{aligned}
& x=\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B} \\
& =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B}=\frac{\sin (A-B)}{\cos A \cos B} \\
& y=\frac{\cos B}{\sin B}-\frac{\cos A}{\sin A} \\
& =\frac{\sin A \cos B-\cos A \sin B}{\sin A \sin B} \\
& =\frac{\sin (A-B)}{\sin A \sin B}\\
& \frac{y}{x}=\frac{\cos \mathrm{A} \cos \mathrm{B}}{\sin \mathrm{A} \sin \mathrm{B}}=\cot \mathrm{A} \cot \mathrm{B} \\
& \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A} \\
& =\frac{\frac{y}{x}+1}{y}=\frac{x+y}{x y} \\
& =\frac{1}{x}+\frac{1}{y} \\
\end{aligned}$

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