How do you prove ${tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5}) = {tan}^{- 1} (\frac{4}{7})$ $tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)$ ?

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How do you prove ${tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5}) = {tan}^{- 1} (\frac{4}{7})$ $tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)$ ?

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Answer 1 · 2016-10-20T21:18:42

See explanation...

Explanation:

For any $α$ $alpha$ and $β$ $beta$ we have:

$tan (α + β) = \frac{tan (α) + tan (β)}{1 - tan (α) tan (β)}$ $tan(alpha+beta) = (tan(alpha)+tan(beta)) / (1-tan(alpha)tan(beta))$

Let $α = {tan}^{- 1} (\frac{1}{3})$ $alpha = tan^(-1)(1/3)$ and $β = {tan}^{- 1} (\frac{1}{5})$ $beta = tan^(-1)(1/5)$

Note that $α \in [0, 1)$ $alpha in [0, 1)$ and $β \in [0, 1)$ $beta in [0, 1)$ , so $α < {tan}^{- 1} = \frac{π}{4}$ $alpha < tan^(-1) = pi/4$ and $β < \frac{π}{4}$ $beta < pi/4$ . Hence $α + β < \frac{π}{2}$ $alpha+beta < pi/2$ is in the range of ${tan}^{- 1}$ $tan^(-1)$

Then we find:

$tan (α + β) = \frac{tan (α) + tan (β)}{1 - tan (α) tan (β)}$ $tan(alpha+beta) = (tan(alpha) + tan(beta))/(1-tan(alpha)tan(beta))$

$tan (α + β) = \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \cdot \frac{1}{5}}$ $color(white)(tan(alpha+beta)) = (1/3 + 1/5)/(1-1/3*1/5)$

$tan (α + β) = \frac{\frac{8}{15}}{\frac{14}{15}}$ $color(white)(tan(alpha+beta)) = (8/15)/(14/15)$

$tan (α + β) = \frac{8}{14}$ $color(white)(tan(alpha+beta)) = 8/14$

$tan (α + β) = \frac{4}{7}$ $color(white)(tan(alpha+beta)) = 4/7$

So:

${tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5}) = α + β = {tan}^{- 1} (\frac{4}{7})$ $tan^(-1)(1/3) + tan^(-1)(1/5) = alpha + beta = tan^(-1)(4/7)$

Answer 2 · 2016-10-20T21:31:56

See the proof below
${tan}^{- 1} (\frac{4}{7}) = {tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5})$ $tan^-1(4/7)=tan^-1(1/3)+tan^-1(1/5)$

Explanation:

Let us have two angles a and b
Then $tan (a + b) = \frac{sin (a + b)}{cos (a + b)} = \frac{sin a cos b + sin b cos a}{cos a cos b - sin a sin b}$ $tan(a+b)=sin(a+b)/cos(a+b)=( sinacosb+sinbcosa)/(cosacosb-sinasinb)$
Divide throughout by $cos a cos b$ $cosacosb$
$tan (a + b) = \frac{(\frac{sin a cos b}{cos a cos b}) + (\frac{sin b cos a}{cos a cos b})}{\frac{cos a cos b}{cos a cos b} - \frac{sin a sin b}{cos a cos b}}$ $tan(a+b)=(((sinacosb)/(cosacosb))+((sinbcosa)/(cosacosb)))/((cosacosb)/(cosacosb)-(sinasinb)/(cosacosb)$
Simplifyng we get
$tan (a + b) = \frac{tan a + tan b}{1 - tan a tan b}$ $tan(a+b)=(tana +tanb)/(1-tanatanb)$
Here we have $tan a = \frac{1}{3}$ $tana=1/3$ and $tan b = \frac{1}{5}$ $tanb=1/5$
so $tan (a + b) = \frac{\frac{1}{3} + \frac{1}{5}}{1 - (\frac{1}{3} \cdot \frac{1}{5})} = \frac{\frac{8}{15}}{\frac{14}{15}} = \frac{8}{14} = \frac{4}{7}$ $tan(a+b)=(1/3+1/5)/(1-(1/3*1/5))=(8/15)/(14/15)=8/14=4/7$
${tan}^{- 1} (\frac{4}{7}) = {tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5})$ $tan^-1(4/7)=tan^-1(1/3)+tan^-1(1/5)$

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How do you prove ${tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5}) = {tan}^{- 1} (\frac{4}{7})$ $tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)$ ?

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How do you prove ${tan}^{- 1} (\frac{1}{3}) + {tan}^{- 1} (\frac{1}{5}) = {tan}^{- 1} (\frac{4}{7})$ $tan^-1(1/3)+tan^-1(1/5)=tan^-1(4/7)$ ?