In any ABC prove the following- i a sinA- b sinB - c sinA-B ii c-b cosA-b-c cosA - cosB- cosC iii a2 sinB-C - b2-c2 sinA iv a-b-a-b - tanA-B-2- tanA-B-2 va cosB-bc cosA-a2-b2

(i) a sin A b sin B =2R(sin^2A sin^2B)=(2Rsin C)(sin (A B))= c sin(A B) (ii) c bcos Ab ccos A = acos Bacos C=cos Bcos C (iii) a^2sin (B C) = 4R^ ...

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Byju's Answer

Standard XII

Mathematics

Conditional Identities

In any ABC ...

Question

In any $△ A B C$ prove the following:
$(i) a s i n A - b s i n B = c s i n (A - B)$
$(i i) \frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}$
$(i i i) a^{2} s i n (B - C) = (b^{2} - c^{2}) s i n A$
$(i v) \frac{(a - b)}{(a + b)} = \frac{tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})}$
$(v) a c o s B - b c c o s A = (a^{2} - b 2)$

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Solution

$(i) a sin A - b sin B = 2 R ({sin}^{2} A - {sin}^{2} B) = (2 R sin C) (sin (A - B)) = c sin (A - B)$

$(i i) \frac{c - b cos A}{b - c cos A} = \frac{a cos B}{a cos C} = \frac{cos B}{cos C}$

$(i i i) a^{2} sin (B - C) = 4 R^{2} {sin}^{2} A sin (B - C)$
$= [4 R^{2} sin A] sin (B + C) . sin (B - C)$ $∵ sin A = sin (B + C)]$

$= [4 R^{2} sin A] [{sin}^{2} B - {sin}^{2} C]$

$= [(2 R sin B)^{2} - (2 R sin C)^{2}] sin A$

$= (b^{2} - c^{2}) sin A$

$(i v) \frac{(a - b)}{(a + b)} = \frac{sin A - sin B}{sin A + sin B} = \frac{2 sin (\frac{A - B}{2}) cos (\frac{A + B}{2})}{2 sin (\frac{A + B}{2}) cos (\frac{A - B}{2})} = \frac{tan (\frac{A + B}{2})}{tan (\frac{A + B}{2})}$

$(v) a cos B - b c cos A = 2 R [sin A cos B - sin B cos A] .2 R sin C$
$= 4 R^{2} sin (A - B) sin (A + B)$

$= 4 R^{2} ({sin}^{2} A - {sin}^{2} B) [∵ a = 2 R sin A]$

$= (a^{2} - b^{2})$

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In any △ABC prove the following:(i)a sinA−b sinB=c sin(A−B)(ii)c−b cosAb−c cosA= cosB cosC(iii)a2 sin(B−C)=(b2−c2) sinA(iv)(a−b)(a+b)=tan(A−B2) tan(A+B2)(v)a cosB−bc cosA=(a2−b2)