Express a trig equation in terms of $k$

Question

The acute angle $x$ radians is such that $\tan x$ = $k$, where $k$ is a positive integer. Express in terms of $k$.

i) $\tan (\pi - x)$
ii) $\tan (\frac{1}{2}\pi - x )$
iii) $\sin x$

I don't understand what the question is stating or asking. I got the answer $\tan \pi - k$ for the first one but that is wrong :(

Edit: Correct answer for the first one is $\tan(\pi - x) = -k$, I don't know how to get there.

Answer 1

Hint: Use $\tan (\alpha +\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$.

Answer 2

1)The $tangent$ function is negative in the second quadrant.

So,$$tan(\pi - x) = -tan(x) = -k$$

2) $tan(\frac{\pi}{2} - x) = cot(x) = \frac{1}{tanx} = \frac{1}{k}$

3)Let us assume a triangle of base(a = 1), height(b = k) w.r.t the angle x (as tanx = b/a = k/1 = k).

So, by Pythagorean theorem, $c = \sqrt{a^2+b^2} = \sqrt{ 1 +k^2}$

$sin x = b/c = \frac{k}{\sqrt{1+k^2}}$

Answer 3

You can use this when you deal trig. Recall that $\tan z=\frac{\sin z}{\cos z}$ and $\sin^2{z}=\frac{1}{1+cotan^2 z}$