Question #82a02

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Question #82a02

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Answer 1 · 2017-10-07T06:01:18

Here X= ${tan}^{- 1} (\frac{1 - {(3)}^{\frac{1}{2}}}{1 + {(3)}^{\frac{1}{2}}})$ $Tan^-1((1-(3)^(1/2)]/(1+(3)^(1/2)))$

Explanation:

There is an identity $tan (x + y)$ $Tan(x+y)$ = $\frac{tan (x) + tan (y)}{1 - tan (x) tan (y)}$ $(Tan(x)+tan(y))/(1-tan(x)tan(y)$

When we apply it to your equation we get
$tan (x + 60) = \frac{tan (x) + tan (60)}{1 - tan (x) tan (60)}$ $tan(x+60)=(tan(x)+tan(60))/(1-tan(x)tan(60)$
As we know $tan (60) = {(3)}^{\frac{1}{2}}$ $tan(60)=(3)^(1/2)$

Therefore $\frac{tan (x) + {(3)}^{\frac{1}{2}}}{1 - {(3)}^{\frac{1}{2}} tan x} = 1$ $(tan(x)+(3)^(1/2))/(1-(3)^(1/2)tanx)=1$

$tan (x) + {(3)}^{\frac{1}{2}} = 1 - {(3)}^{\frac{1}{2}} tan (x)$ $tan(x)+(3)^(1/2)=1-(3)^(1/2)tan(x)$

$tan (x) + {(3)}^{\frac{1}{2}} tan (x) = 1 - {(3)}^{\frac{1}{2}}$ $tan(x)+(3)^(1/2)tan(x)=1-(3)^(1/2)$

$tan (x) (1 + {(3)}^{\frac{1}{2}}) = 1 - {(3)}^{\frac{1}{2}}$ $tan(x)(1+(3)^(1/2))=1-(3)^(1/2)$

$tan (x) = \frac{1 - {(3)}^{\frac{1}{2}}}{1 + {(3)}^{\frac{1}{2}}}$ $tan(x)=(1-(3)^(1/2))/(1+(3)^(1/2)$

Therefore $x = {tan}^{- 1} (\frac{1 - {(3)}^{\frac{1}{2}}}{1 + {(3)}^{\frac{1}{2}}})$ $x=tan^-1((1-(3)^(1/2))/(1+(3)^(1/2)))$

Answer 2 · 2017-10-07T06:16:09

A different approach...
See the answer below...

Explanation:

$tan (x + 60) = 1$ $tan(x+60)=1$
$\Rightarrow tan (x + 60) = tan 45$ $=>tan(x+60)=tan45$ (mainly)

We might take further higher value...
$x + 60 = 45$ $x+60=45$
$\Rightarrow x = - 15$ $=>x=-15$

Hope this helps...

It is same as ${tan}^{- 1} (\frac{1 - \sqrt{3}}{1 + \sqrt{3}})$ $tan^-1((1-sqrt3)/(1+sqrt3))$ ...

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Question #82a02

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