In any - ABC- prove that B - C - B - C -tan1-2 B - C -tan1-2 B - C

By the sine rule, we have a/sin A = b/sin B = c/sin C = k (say) ⇒ a = k sin A, b = k sin B and c k sin C ∴ LHS = (b c)/b + c = k sin B k sin C/k sin B + ...

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Byju's Answer

Standard XI

Mathematics

Sin2a and Cos2a in Terms of Tana

In any Δ ABC,...

Question

In any $Δ A B C,$ prove that

$\frac{(B - C)}{(B + C)} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$

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Solution

By the sine rule, we have

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = k (s a y)$

$\Rightarrow a = k s i n A, b = k s i n B a n d c - k s i n C$

$∴ L H S = \frac{(b - c)}{b + c} = \frac{k s i n B - k s i n C}{k s i n B + k s i n C} = \frac{k (s i n B - s i n C)}{k (s i n B + s i n C)}$

= $\frac{(s i n B - s i n C)}{(s i n B + s i n C)} = \frac{2 c o s \frac{(B + C)}{2} s i n \frac{(B - C)}{2}}{2 s i n \frac{B + C}{2} c o s \frac{(B - C)}{2}}$

= $\frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)} = R H S .$

Hence, $\frac{(b - c)}{b + c} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)} .$

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Similar questions

In any $Δ A B C,$ prove that

$\frac{(b - c)}{b + c} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$

In $Δ A B C,$ prove that

$\frac{(b - c)}{b + c} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$

Q. Prove that

\frac{c}{a + b} = \frac{1 - tan \frac{1}{2} A tan \frac{1}{2} B}{1 + tan \frac{1}{2} A tan \frac{1}{2} B}

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\frac{tan \frac{1}{2} A}{(a - b) (a - c)} + \frac{tan \frac{1}{2} B}{(b - c) (b - a)} + \frac{tan \frac{1}{2} C}{(c - a) (c - b)} = \frac{1}{S}

Q. Prove that:

1 - tan \frac{1}{2} A tan \frac{1}{2} B = \frac{2 c}{(a + b + c)}

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In any ΔABC, prove that (B−C)(B+C)=tan 12(B−C)tan 12(B+C)

In any $Δ A B C,$ prove that

$\frac{(B - C)}{(B + C)} = \frac{t a n \frac{1}{2} (B - C)}{t a n \frac{1}{2} (B + C)}$