If in a -ABC- a tan A - b tan B - a - b tan A - B2 - then

The correct option is B A = B a(tan A tan(A+B/2))=b(tan(A+B/2) tan B) ⇒ a(sinA.cos(A+B/2) sin (A+B/2).cosA)cos A=b(sin(A+B/2)cos A sin Acos (A+B/2))c ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Half Angles

If in a ΔAB...

Question

If in a $Δ$ ABC, $a t a n A + b t a n B = (a + b) t a n (\frac{A + B}{2})$ , then

A = B

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A = - B

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A = 2 B

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B = 2A

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Solution

The correct option is B A = - B
$a (t a n A - t a n (\frac{A + B}{2})) = b (t a n (\frac{A + B}{2}) - t a n B) \Rightarrow a \frac{(s i n A . c o s (\frac{A + B}{2}) - s i n (\frac{A + B}{2}) . c o s A)}{c o s A} = b \frac{(s i n (\frac{A + B}{2}) c o s A - s i n A c o s (\frac{A + B}{2}))}{c o s B}$
$\Rightarrow \frac{a}{c o s A} = \frac{- b}{c o s B}$ .........Eq(1)

By Sine Rule
$\frac{a}{s i n A} = \frac{b}{s i n B} = k$
$∴ a = k s i n A, b = k s i n B$
Putting the above values in Eq(1),we get
$\frac{k s i n A}{c o s A} = \frac{- k s i n B}{c o s B}$
$\Rightarrow t a n A = - t a n B$
$\Rightarrow t a n A = t a n (- B)$
$∴ A = - B$

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