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Question Number 210261 by klipto last updated on 04/Aug/24

$$\mathbf{show}\mathbf{that}$$$$\frac{\mathbf{sinAcosA}-\mathbf{sinBcosB}}{{\mathbf{cos}}^2\mathbf{A}-{\mathbf{sin}}^2\mathbf{B}}=\mathbf{tan}(\mathbf{A}-\mathbf{B})$$

Answered by efronzo1 last updated on 04/Aug/24

$$\frac{\sin \mathrm{A}\cos \mathrm{A}-\sin \mathrm{B}\cos \mathrm{B}}{\cos {}^2\mathrm{A}-\sin {}^2\mathrm{B}}\stackrel{?}{=}\tan (\mathrm{A}-\mathrm{B})$$$$\underbrace{⋐}$$

Answered by A5T last updated on 04/Aug/24

$$tan(A-B)=\frac{sin(A-B)}{cos(A-B)}=\frac{sinAcosA-sinBcosB}{{cos}^{2}A-{sin}^{2}B}$$

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