Question
If $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}},$ then write the value of $\tan\text{x}\tan\text{y}.$

Answer

We have, $\frac{\cos(\text{x - y})}{\cos(\text{x+y})}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\frac{\cos\text{x}\cos\text{y}+\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y}}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\frac{\frac{\cos\text{x}\cos\text{y}+}{\cos\text{x}\cos\text{y}}+\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}{\frac{\cos\text{x}\cos\text{y}}{\cos\text{x}\cos\text{y}}-\frac{\sin\text{x}\sin\text{y}}{\cos\text{x}\cos\text{y}}}=\frac{\text{m}}{\text{n}}$ $[$Dividing and multiplying numerator and denominator by $\cos\text{x}\cos\text{y}]$ $\Rightarrow\frac{1+\tan\text{x}\tan\text{y}}{1-\tan\text{x}\tan\text{y}}=\frac{\text{m}}{\text{n}}$ $\Rightarrow\text{n+n}\tan\text{x}\tan\text{y}=\text{m - m}\tan\text{x}\tan\text{y}$ $\Rightarrow\tan\text{x}\tan\text{y}(\text{n+m})=\text{m - n}$ $\Rightarrow\tan\text{x}\tan\text{y}=\frac{\text{m - n}}{\text{m+n}}$

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