Question #2533c

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Answer 1 · 2018-02-14T03:13:48

$- 1$ $-1$

This is based on the double angle formula for $tan x$ $tanx$

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

x, in your problem, is $\frac{7 π}{8}$ $(7pi)/8$ , so your problem is really looking for :

$tan (2 (\frac{7 π}{8})) = \frac{2 tan (\frac{7 π}{8})}{1 - {tan}^{2} (\frac{7 π}{8})}$ $tan(2((7pi)/8))=(2tan((7pi)/8))/(1-tan^2((7pi)/8))$

$tan (2 (\frac{7 π}{8})) = tan (\frac{14 π}{8}) = tan (\frac{7 π}{4})$ $tan(2((7pi)/8))=tan((14pi)/8)=tan((7pi)/4)$

$tan (\frac{7 π}{4}) = - 1$ $tan((7pi)/4)=-1$

Answer 2 · 2018-02-14T03:20:33

See the explanation

$\frac{2 tan (\frac{7 π}{8})}{1 - {tan}^{2} (\frac{7 π}{8})}$ $[2tan((7pi)/8)]/(1 - tan^2((7pi)/8)$

It is in the form of

$\frac{2 tan x}{1 - {tan}^{2} x}$ $(2tanx)/(1 - tan^2x)$

Which is an identity and is equal to $tan 2 x$ $tan2x$ .

So, $\frac{2 tan (\frac{7 π}{8})}{1 - {tan}^{2} (\frac{7 π}{8})}$ $[2tan((7pi)/8)]/(1 - tan^2((7pi)/8)$

$= tan (2 \cdot \frac{7 π}{8})$ $= tan(2*(7pi)/8)$

$= tan (\frac{7 π}{4})$ $= tan((7pi)/4)$

Now put $pi = 180°$

We get , $tan(7*180/4)$

$= tan(7*45)$

$= tan (315°)$

Now you can use the tables to find the value of $tan (315°)$ which is equal to -1.