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Question

# Find:(a) $\tan (\alpha+\beta)$ and(b) $\tan (\alpha-\beta)$(c) $\tan \alpha=2$(d) $\tan \beta=-\frac{1}{3}$

Solution

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Step 1
1 of 2

a) Use sum formula:

$\tan(\alpha+\beta)=\\\\ =\dfrac{2-\dfrac{1}{3}}{1-2\cdot (-\dfrac{1}{3})}=\dfrac{\dfrac{5}{3}}{\dfrac{5}{3}}=1$

b) Use difference formula:

$\tan(\alpha-\beta)=\\\\ =\dfrac{2-(-\dfrac{1}{3})}{1+2\cdot (-\dfrac{1}{3})}=\dfrac{\dfrac{7}{3}}{\dfrac{1}{3}}=\dfrac{7}{1} =7$

Sum formula for Tangent:$\\\\$$\tan(\alpha+\beta)=$\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\\\\\\ Difference formula for Tangent:$ \tan(\alpha-\beta)=$$\dfrac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}$

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