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Question

Find:

(a) tan(α+β)\tan (\alpha+\beta) and

(b) tan(αβ)\tan (\alpha-\beta)

(c) tanα=2\tan \alpha=2

(d) tanβ=13\tan \beta=-\frac{1}{3}

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 2

a) Use sum formula:

tan(α+β)==21312(13)=5353=1\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(αβ)==2(13)1+2(13)=7313=71=7\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)=tanα+tanβ1tanαtanβDifferenceformulaforTangent:\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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