Question

Please someone help me to prove this. ​

Answer 1

We know that,

`\dag\bf\:sin^2A=(1-cos2A)/(2)`

`\dag\bf\:sin2A=2sinA\:cosA`

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Now, Let's solve !

`\leadsto\:\bf(sin^2A-sin^2B)/(sinA\:cosA-sinB\:cosB)`

`\leadsto\:\sf((1-cos2A)/(2)-(1-cos2B)/(2))/((2sinA\:cosA)/(2)-(2sinB\:cosB)/(2))`

`\leadsto\:\sf(1-cos2A-1+cos2B)/(sin2A-sin2B)`

`\leadsto\:\sf(2sin(2A+2B)/(2)\:sin(2A-2B)/(2))/(2sin(2A-2B)/(2)\:cos(2A+2B)/(2))`

`\leadsto\:\sf(sin(A+B))/(cos(A+B))`

`\leadsto\:\bf{tan(A+B)}`

Answer 2

Answer: see proof below

Explanation:

Use the Power Reducing Identity: sin² Ф = (1 - cos 2Ф)/2

Use the Double Angle Identity: sin 2Ф = 2 sin Ф · cos Ф

Use the following Sum to Product Identities:

`\sin x - \sin y = 2\cos \bigg((x+y)/(2)\bigg)\sin \bigg((x-y)/(2)\bigg)\\\\\\\cos x - \cos y = -2\sin \bigg((x+y)/(2)\bigg)\sin \bigg((x-y)/(2)\bigg)`

Proof LHS → RHS

`\text{LHS:}\qquad \qquad \qquad (\sin^2A-\sin^2B)/(\sin A\cos A-\sin B \cos B)`

`\text{Power Reducing:}\qquad (\bigg((1-\cos 2A)/(2)\bigg)-\bigg((1-\cos 2B)/(2)\bigg))/(\sin A \cos A-\sin B\cos B)`

`\text{Half-Angle:}\qquad \qquad (\bigg((1-\cos 2A)/(2)\bigg)-\bigg((1-\cos 2B)/(2)\bigg))/((1)/(2)\bigg(\sin 2A-\sin 2B\bigg))`

`\text{Simplify:}\qquad \qquad (1-\cos 2A-1+\cos 2B)/(\sin 2A-\sin 2B)\\\\\\.\qquad \qquad \qquad =(-\cos 2A+\cos 2B)/(\sin 2A - \sin 2B)\\\\\\.\qquad \qquad \qquad =(\cos 2B-\cos 2A)/(\sin 2A-\sin 2B)`

`\text{Sum to Product:}\qquad \qquad (-2\sin \bigg((2B+2A)/(2)\bigg)\sin \bigg((2B-2A)/(2)\bigg))/(2\cos \bigg((2A+2B)/(2)\bigg)\sin \bigg((2A-2B)/(2)\bigg))`

`\text{Simplify:}\qquad \qquad (-2\sin (A + B)\cdot \sin (-[A - B]))/(2\cos (A + B) \cdot \sin (A - B))`

`\text{Co-function:}\qquad \qquad (2\sin (A + B)\cdot \sin (A - B))/(2\cos (A + B) \cdot \sin (A - B))`

`\text{Simplify:}\qquad \qquad \quad (\cos (A+B))/(\sin (A+B))\\\\\\.\qquad \qquad \qquad \quad =\tan (A+B)`

LHS = RHS: tan (A + B) = tan (A + B)
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