In any triangle ABC - tan A - 2 -tan B - L -tan A - t -tan BA- a-b-a-b-cB- d-a-bC- a-CD- C-a-b

The correct option is C a b/cUsing formula of incentre; r = (s a)tanA/2; ⇒ tanA/2=r/s a Similarly; tanB/2=r/s b tan A/2 tan B/2/tan A/2+tan B/2=r/s a r/s b/r/s ...

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Standard X

Mathematics

Trigonometric Identities

In any triang...

Question

In any triangle ABC, $\frac{t a n \frac{A}{2} - t a n \frac{B}{2}}{t a n \frac{A}{2} + t a n \frac{B}{2}}$

\frac{a - b}{a + b}

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\frac{a - b}{c}

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\frac{a - b}{a + b + c}

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\frac{c}{a + b}

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Solution

The correct option is B $\frac{a - b}{c}$
Using formula of incentre; $r = (s - a) t a n \frac{A}{2};$
$\Rightarrow t a n \frac{A}{2} = \frac{r}{s - a}$
Similarly; $t a n \frac{B}{2} = \frac{r}{s - b}$
$\frac{t a n \frac{A}{2} - t a n \frac{B}{2}}{t a n \frac{A}{2} + t a n \frac{B}{2}} = \frac{\frac{r}{s - a} - \frac{r}{s - b}}{\frac{r}{s - a} + \frac{r}{s - b}}$
$= \frac{(s - b) - (s - a)}{(s - b) + (s - a)}$
$= \frac{a - b}{2 s - a - b}$
$= \frac{a - b}{c}$ $[Since s = \frac{a + b + c}{2}]$

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In any triangle ABC, tanA2−tanB2tanA2+tanB2

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In any triangle ABC, $\frac{t a n \frac{A}{2} - t a n \frac{B}{2}}{t a n \frac{A}{2} + t a n \frac{B}{2}}$