If A - B = x and B - A = y, then what is (A-B) equal to?

Correct Answer is : 1x+1y1/x + 1/yx1+y1

Correct Answer is : 1x+1y1/x + 1/yx1+y1

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UPSC NDA II 2011 Mathematics Paper

Question : 45 of 120

Marks: +1, -0

\tan A - \tan B = x

\cot B - \cot A = y

\cot(A-B)

equal to?

Solution:

\because \;\; \tan A - \tan B = x

\Rightarrow \;\; \frac{\sin A}{\cos A} - \frac{\sin B}{\cos B} = x

\Rightarrow \;\; \frac{\sin(A-B)}{\cos A \cdot \cos B} = x

\Rightarrow \;\; \sin(A-B) = x \cos A \cdot \cos B \;\; \ldots (i)

\cot B - \cot A = y

\frac{\cos B}{\sin B} - \frac{\cos A}{\sin A} = y

\Rightarrow \;\; \sin(A-B) = y \sin A \cdot \sin B \;\; \ldots

(ii)

From (i) and

(

) x \cos A \cdot \cos B = y \sin A \cdot \sin B

\cot(A-B) = \frac{\cos(A-B)}{\sin(A-B)}

= \frac{\cos A \cdot \cos B + \sin A \cdot \sin B}{x \cos A \cdot \cos B}

\left[ \because x \cos A \cdot \cos B = y \sin A \cdot \sin B \right]

= \frac{1}{x} + \frac{\sin A \cdot \sin B}{x \cos A \cdot \cos B}

= \frac{1}{x} + \frac{\sin A \cdot \sin B}{y \sin A \cdot \sin B}

= \frac{1}{x} + \frac{1}{y}

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