If $tan θ = \frac{5}{6}$ $tan theta = 5/6$ and $tan ϕ = \frac{1}{11}$ $tan phi= 1/11$ then find $tan (θ + ϕ)$ $tan (theta + phi)$ hence show that $(θ + ϕ) = \frac{π e}{4}$ $(theta + phi) = (pie)/4$ ?

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If $tan θ = \frac{5}{6}$ $tan theta = 5/6$ and $tan ϕ = \frac{1}{11}$ $tan phi= 1/11$ then find $tan (θ + ϕ)$ $tan (theta + phi)$ hence show that $(θ + ϕ) = \frac{π e}{4}$ $(theta + phi) = (pie)/4$ ?

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Answer 1 · 2018-05-21T16:26:34

Proved as described below.

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Explanation:

$tan θ = \frac{5}{6}$ $tantheta=5/6$

$tan ϕ = \frac{1}{11}$ $tanphi=1/11$

$tan (θ + ϕ) = \frac{tan θ + tan ϕ}{1 - tan θ tan ϕ}$ $tan(theta+phi)=(tantheta+tanphi)/(1-tanthetatanphi)$

$= \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} \times \frac{1}{11}}$ $=(5/6+1/11)/(1-5/6xx1/11)$

$= \frac{\frac{5 \times 11 + 1 \times 6}{6 \times 11}}{\frac{6 \times 11 - 5 \times 1}{1 \times 6 \times 11}}$ $=((5xx11+1xx6)/(6xx11))/((6xx11-5xx1)/(1xx6xx11))$

$= \frac{\frac{55 + 6}{66}}{\frac{66 - 5}{66}}$ $=((55+6)/66)/((66-5)/66)$

$= \frac{61}{61} = 1$ $=61/61=1$

$tan (θ + ϕ) = 1$ $tan(theta+phi)=1$

$tan (\frac{π}{4}) = 1$ $tan(pi/4)=1$

$θ + ϕ = \frac{π}{4}$ $theta+phi=pi/4$

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If $tan θ = \frac{5}{6}$ $tan theta = 5/6$ and $tan ϕ = \frac{1}{11}$ $tan phi= 1/11$ then find $tan (θ + ϕ)$ $tan (theta + phi)$ hence show that $(θ + ϕ) = \frac{π e}{4}$ $(theta + phi) = (pie)/4$ ?

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If $tan θ = \frac{5}{6}$ $tan theta = 5/6$ and $tan ϕ = \frac{1}{11}$ $tan phi= 1/11$ then find $tan (θ + ϕ)$ $tan (theta + phi)$ hence show that $(θ + ϕ) = \frac{π e}{4}$ $(theta + phi) = (pie)/4$ ?