Question

Verify tan(α+β) = [tan(α)+tan(β)] / [1 + tan(α)tan(β)]

Answer 1

Answer:

Proved

Step-by-step explanation:

To Prove: $tan(\alpha+\beta) =\dfrac{tan(\alpha)+tan(\beta)}{1 + tan(\alpha)tan(\beta)}$

Proof:

Now: $tan \theta =\dfrac{sin\theta }{cos \theta}$

Therefore:

$tan (\alpha+\beta)=\dfrac{ sin (\alpha+\beta)}{cos (\alpha+\beta) }$

Applying these angle sum formula

$sin (\alpha+\beta)=sin \alpha cos \alpha + sin \beta cos \beta\\cos (\alpha+\beta)=cos \alpha cos \beta - sin \alpha sin \beta$

$tan (\alpha+\beta)=\dfrac{ sin \alpha cos \alpha + sin \beta cos \beta}{cos \alpha cos \beta - sin \alpha sin \beta }$

Divide all through by $cos \alpha cos \beta$

$tan (\alpha+\beta)=\dfrac{ (sin \alpha cos \alpha)/(cos \alpha cos \beta) + (sin \beta cos \beta)/(cos \alpha cos \beta)}{(cos \alpha cos \beta)/(cos \alpha cos \beta) - (sin \alpha sin \beta)/(cos \alpha cos \beta) }\\\\tan (\alpha+\beta)=\dfrac{\frac{sin \alpha}{cos \alpha}+\frac{sin \beta}{cos \beta} }{1-tan \alpha tan \beta} \\ Therefore:\\\\tan (\alpha+\beta)=\dfrac{tan \alpha+tan \beta}{1-tan \alpha tan \beta}$

=sin \alpha cos \beta + cos \alpha sin \beta/cos \alpha cos \beta/cos \alpha cos \beta- sin \alpha sin \beta/cos \alpha cos \beta  

=sin \alpha/cos \alpha + sin \beta/cos \beta/1-tan \alpha tan \beta  

=tan A + tan B/1-tan A tan B