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Question Number 163153 by MathsFan last updated on 04/Jan/22

$$showthat$$$$\frac{cos(x-y)}{cos(x+y)}=\frac{1+tanxtany}{1-tanxtany}$$

Answered by cortano1 last updated on 04/Jan/22

$$RHS:\frac{\cos x\cos y+\sin x\sin y}{\cos x\cos y-\sin x\sin y}$$$$=\frac{\cos (x-y)}{\cos (x+y)}$$

Answered by Rasheed.Sindhi last updated on 04/Jan/22

$$\frac{cos(x-y)}{cos(x+y)}=\frac{1+tanxtany}{1-tanxtany}$$$$\mathrm{LHS}:\frac{cos(x-y)}{cos(x+y)}$$$$=\frac{cosxcosy+sinxsiny}{cosxcosy+sinxsiny}$$$$=\frac{\frac{cosxcosy+sinxsiny}{cosxcosy}}{\frac{cosxcosy-sinxsiny}{cosxcosy}}$$$$=\frac{1+\frac{sinxsiny}{cosxcosy}}{1-\frac{sinxsiny}{cosxcosy}}=\frac{1+\frac{sinx}{cosx}\cdot \frac{siny}{cosy}}{1-\frac{sinx}{cosx}\cdot \frac{siny}{cosy}}$$$$=\frac{1+tanxtany}{1-tanxtany}=\mathrm{RHS}$$

Commented by MathsFan last updated on 04/Jan/22

$$merci$$

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