Question

Please someone help me to prove this. ​

Answer 1

Answer:  see proof below

Step-by-step 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:

$\sin x - \sin y = 2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x - \cos y = -2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

Proof LHS →  RHS

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin^2A-\sin^2B}{\sin A\cos A-\sin B \cos B}$

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

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

$\text{Simplify:}\qquad \qquad \dfrac{1-\cos 2A-1+\cos 2B}{\sin 2A-\sin 2B}\\\\\\.\qquad \qquad \qquad =\dfrac{-\cos 2A+\cos 2B}{\sin 2A - \sin 2B}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 2B-\cos 2A}{\sin 2A-\sin 2B}$

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

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

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

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

LHS = RHS:    tan (A + B) = tan (A + B)    $\checkmark$

Answer 2

Answer :

We know that,

$\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}$

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

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Now, Let's solve !

$\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}$

$\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}$

$\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}$

$\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}$

$\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}$

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