-sin A-sin B-sin A-sin B-a-b-2 -2 b-a-b

√(sin A) √(sin B)/√(sin A)+√(sin B)=a+b 2√(ab)/a b RHS a+b 2√(ab)/a b =(√(a) )^2+(√(b) )^2 2√(ab)/(√(a) )^2 (√(b) )^2 =(√(a) √(b) )^2/(√(a) )^2 (√(b) )^2 =( ...

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Byju's Answer

Standard XII

Mathematics

Equations of the Form acosθ+bsinθ = c

√sin A-√sin B...

Question

$\frac{\sqrt{s i n A} - \sqrt{s i n B}}{\sqrt{s i n A} + \sqrt{s i n B}} = \frac{a + b - 2 \sqrt{a b}}{a - b}$

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Solution

$\frac{\sqrt{s i n A} - \sqrt{s i n B}}{\sqrt{s i n A} + \sqrt{s i n B}} = \frac{a + b - 2 \sqrt{a b}}{a - b} R H S \frac{a + b - 2 \sqrt{a b}}{a - b} = \frac{{(\sqrt{a})}^{2} + {(\sqrt{b})}^{2} - 2 \sqrt{a b}}{{(\sqrt{a})}^{2} - {(\sqrt{b})}^{2}} = \frac{{(\sqrt{a} - \sqrt{b})}^{2}}{{(\sqrt{a})}^{2} - {(\sqrt{b})}^{2}} = \frac{(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})} = \frac{(\sqrt{k s i n A} - \sqrt{k s i n B})}{(\sqrt{k s i n A} + \sqrt{k s i n B})} = \frac{(\sqrt{s i n A} - \sqrt{s i n B})}{(\sqrt{s i n A} + \sqrt{s i n B})} [taking k common and cancelling them] = L H S Hence proved.$

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Similar questions

\frac{sin A + sin B + sin C}{sin A + sin B - sin C} = cot \frac{A}{2} cot \frac{B}{2}

Q. In triangle

A B C, \frac{sin A + sin B + sin C}{sin A + sin B - sin C}

is equal to

Q. If

A + B + C = π

, prove that

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

Q. In any

△ A B C

, prove that

\frac{\sqrt{sin A} - \sqrt{sin B}}{\sqrt{sin A} + \sqrt{sin B}} = \frac{a + b - 2 \sqrt{a b}}{a - b}

(Hint:

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

Prove that SinA+sinB=sin(A+B)cos(A+B)/2

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√sin A − √sin B√sin A + √sin B=a+b−2√aba−b

$\frac{\sqrt{s i n A} - \sqrt{s i n B}}{\sqrt{s i n A} + \sqrt{s i n B}} = \frac{a + b - 2 \sqrt{a b}}{a - b}$