If tan A -tan B - x and B - A - y- where x - y -0- then A - B is- 1- x -1- y- 1- x -1-1- y- 1- y -1- xD- 1- xy -1- y

The correct option is B 1x+ 1yWe have nbsp;tan A tan B=x and B A=y Now, y=tan A tan Btan Atan B ⇒ y= xtan Atan B ⇒ nbsp; A B= yx Now, (A B)= A B+1 B A (A ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

If tan A -tan...

Question

If $tan A - tan B = x$ and $cot B - cot A = y$ , where $x, y \neq 0$ , then $cot (A - B)$ is

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

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

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

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

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Solution

The correct option is B $\frac{1}{x} + \frac{1}{y}$
We have $tan A - tan B = x$ and $cot B - cot A = y$
Now, $y = \frac{tan A - tan B}{tan A tan B}$
$\Rightarrow y = \frac{x}{tan A tan B} \Rightarrow cot A cot B = \frac{y}{x}$
Now,
$cot (A - B) = \frac{cot A cot B + 1}{cot B - cot A}$
$cot (A - B) = \frac{\frac{y}{x} + 1}{y} = \frac{x + y}{x y} = \frac{1}{x} + \frac{1}{y}$

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Trigonometric Ratios of Compound Angles

Standard XII Mathematics

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If tanA−tanB=x and cotB−cotA=y, where x,y≠0, then cot(A−B) is

The correct option is B 1x+1yWe have tanA−tanB=x and cotB−cotA=y Now, y=tanA−tanBtanAtanB ⇒y=xtanAtanB⇒cotAcotB=yx Now, cot(A−B)=cotAcotB+1cotB−cotA cot(A−B)=yx+1y =x+yxy =1x+1y

If $tan A - tan B = x$ and $cot B - cot A = y$ , where $x, y \neq 0$ , then $cot (A - B)$ is