In - 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 XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

In Δ ABC, pro...

Question

In $Δ A B C,$ prove that

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

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Solution

By the sine rule, we have

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

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

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

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

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

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

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

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}

Q. Prove that

\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)}

Q. In a triangle ABC if

t a n \frac{1}{2} A = \frac{5}{6}

and

t a n \frac{1}{2} B = \frac{20}{37}

find

t a n \frac{1}{2} C,

and prove that in this triangle

a + c = 2 b .

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In ΔABC, prove that (b−c)b+c=tan12(B−C)tan12(B+C)

In $Δ A B C,$ prove that

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