MHT CET · Maths · Trigonometric Equations
If \(\mathrm{A}\gt\mathrm{B}\) and \(\tan \mathrm{A}-\tan \mathrm{B}=x\) and \(\cot \mathrm{B}-\cot \mathrm{A}=y\), then \(\cot (\mathrm{A}-\mathrm{B})=\)
- A \(\frac{1}{y}-\frac{1}{x}\)
- B \(\frac{1}{x}-\frac{1}{y}\)
- C \(\frac{1}{x}+\frac{1}{y}\)
- D \(\frac{x y}{x-y}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{x}+\frac{1}{y}\)
Step-by-step Solution
Detailed explanation
Given, \(\tan \mathrm{A}-\tan \mathrm{B}=x\)
\(\cot \mathrm{B}-\cot \mathrm{A}=y\)
\(\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y \)
\(\Rightarrow \frac{\tan A-\tan B}{\tan A \cdot \tan B}=y \)
\(\Rightarrow \tan A \cdot \tan B=\frac{x}{y}...(i)\)
\(\text {Now, } \cot (A-B) =\frac{1}{\tan (A-B)} \)
\(=\frac{1+\tan A \cdot \tan B}{\tan A-\tan B} \)
\(=\frac{1+\frac{x}{y}}{x} \)
\(=\frac{y+x}{x y} \)
\(=\frac{1}{x}+\frac{1}{y}\)
\(\cot \mathrm{B}-\cot \mathrm{A}=y\)
\(\Rightarrow \frac{1}{\tan B}-\frac{1}{\tan A}=y \)
\(\Rightarrow \frac{\tan A-\tan B}{\tan A \cdot \tan B}=y \)
\(\Rightarrow \tan A \cdot \tan B=\frac{x}{y}...(i)\)
\(\text {Now, } \cot (A-B) =\frac{1}{\tan (A-B)} \)
\(=\frac{1+\tan A \cdot \tan B}{\tan A-\tan B} \)
\(=\frac{1+\frac{x}{y}}{x} \)
\(=\frac{y+x}{x y} \)
\(=\frac{1}{x}+\frac{1}{y}\)
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