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KEAM 2016 Math Paper

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Question : 46 of 120

Marks: +1, -0

If

\tan A - \tan B = x

\cot B - \cot y,

\cot(A+B)

is

$\frac{1}{x-y}$
$\frac{1}{x+y}$
$\frac{1}{x} + y$
$\frac{1}{x} - \frac{1}{y}$
$\frac{1}{x} + \frac{1}{y}$

Solution:

Given,

\tan A - \tan B = x

.....(i)

and

\cot B - \cot A = y

.....(ii)

From Eq. (ii),

\cot B - \cot A = y,

we get

\Rightarrow \frac{1}{\tan B} - \frac{1}{\tan A} = y

\Rightarrow \frac{\tan A - \tan B}{\tan B \tan A} = y

\Rightarrow \frac{x}{\tan B \tan A} = y

[from Eq. (i)]

\therefore \cot B \cot A = \frac{y}{x}

....(iii)

\therefore \cot(A-B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}

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

[ from Eq. (ii) and Eq. (iii) ]

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

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