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Question Number 132081 by liberty last updated on 11/Feb/21

$$\mathrm{Prove}\tan (\mathrm{A}+\mathrm{B})-\tan \mathrm{A}=\frac{\sin \mathrm{B}}{\cos \mathrm{A}\cos (\mathrm{A}+\mathrm{B})}$$

Answered by rs4089 last updated on 11/Feb/21

$$tan(A+B)-tanA$$$$\frac{sin(A+B)}{cos(A+B)}-\frac{sinA}{cosA}$$$$\frac{sin(A+B).cosA-sinA.cos(A+B)}{cos(A+B).cosA}$$$$\frac{sin\{(A+B)-\left(A\right)\}}{cos(A+B).cosA}$$$$\frac{sinB}{cos(A+B).cosA}$$

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