If \( \Large \tan A - \tan B = x\ and\ \cot B - \cot A = y,\) then\( \Large  \cot \left(A-B\right) \) is equal to:


A) \( \Large \frac{1}{x}+y \)

B) \( \Large \frac{1}{xy} \)

C) \( \Large \frac{1}{x}-\frac{1}{y} \)

D) \( \Large \frac{1}{x}+\frac{1}{y} \)

Correct Answer:
D) \( \Large \frac{1}{x}+\frac{1}{y} \)

Description for Correct answer:

Given that

\( \Large \tan A - \tan B = x \) ...(i)

and \( \Large \cot B - \cot A = y \) ...(ii)

Now, \( \Large \cot \left(A-B\right) = \frac{1}{\tan \left(A-B\right) } \)

= \( \Large \frac{1+\tan A \tan B}{\tan A - \tan B} \)

= \( \Large \frac{1}{\tan A - \tan B} + \frac{\tan A \tan B}{\tan A - \tan B} \)

= \( \Large \frac{1}{x}+\frac{1}{y} \) [from I and II]


Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio








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