# Trigonometry; $\tan (90^{\circ} + \theta)$?

Q: If $\sin(\theta) = k$ and $\theta$ is obtuse, find an expression for $\tan(\theta + 90^{\circ})$

A: $\frac{\sqrt{(1 - k^2)}}{k}$

I tried drawing the triangle in the second quadrant then flipping it into the third quadrant and the answer was correct but i was wondering whether there was a better method to do it ?

• I think drawing triangles in circles is a great way to do it. Pay attention to you signs, though... if $90<90+\theta<180$, then $\tan(90+\theta) < 0$ – Doug M Jul 15 '16 at 0:24

Just keep track of the signs of the trig ratios and it's easy.

You need to find $\displaystyle \tan{(90^{\circ} + \theta)}$, which is identically equal to $\displaystyle -\cot \theta = -\frac{\cos\theta}{\sin\theta}$.

Since this is a second quadrant angle, you take the negative root for cosine, i.e. $\displaystyle \cos\theta = -\sqrt{1-\sin^2 \theta} = -\sqrt{1-K^2}$

Putting it together, you get:

$$\tan{(90^{\circ} + \theta)} = (-)\frac{-\sqrt{1-K^2}}{K} = \frac{\sqrt{1-K^2}}{K}$$

• sorry this might be a silly question but why does tan(90 + θ) = −cotθ? – kjhg Jul 15 '16 at 0:54
• One way is to use the angle sum formula, but you run into a bit of an issue since $\tan 90^{\circ}$ is undefined (infinite). You can treat it as the limit: $\lim_{\alpha \to 90^{\circ}}\frac{\tan \alpha + \tan \theta }{1 - \tan\alpha \tan \theta}$, which can be shown to be $-\cot \theta$. The easy way to remember it is to recognise that adding or subtracting odd multiples of a right angle to the argument of a tangent changes it to a cotangent, and all you have to worry about is the sign. (cont'd) – Deepak Jul 15 '16 at 1:23
• (cont'd) In this case, if you visualise a first quadrant angle (as a general simple example) and add a right angle to it, you put it into the second quadrant, and you change the sign of the tangent (and therefore cotangent) to negative. This is an easy way to "see" what happens to the signs of various trig ratios when you add or subtract special angles to the arguments. Adding/subtracting an even multiple of a right angle preserves the ratio, but doing the same to an odd multiple of a right angle changes a sine to a cosine (and vice versa) and a tangent to a cotangent. Then consider the sign. – Deepak Jul 15 '16 at 1:27

Write $$\tan(\theta + 90^{\circ}) = \frac{\sin(\theta+90^{\circ})}{\cos(\theta+90^{\circ})}.$$ Now notice that $\sin(\theta+90^{\circ}) = \cos(\theta)$ and that $\cos(\theta+90^{\circ}) = -\sin(\theta)$. To prove these equalities, you can draw triangles or use the formulas $$\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)$$ and $$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B).$$ Since $\sin(90^{\circ})=1$ and $\cos(90^{\circ}) = 0$, we have $$\sin(\theta+90^{\circ}) = \sin(\theta)\cos(90^{\circ}) + \cos(\theta)\sin(90^{\circ}) = \cos(\theta)$$ and $$\cos(\theta+90^{circ}) = \cos(\theta)\cos(90^{\circ})-\sin(\theta)\sin(90^{\circ}) = - \sin(\theta).$$

Therefore, we have $$\tan(\theta+90^{\circ}) = -\frac{\cos(\theta)}{\sin(\theta)}.$$ Since $\cos(\theta) = \pm \sqrt{1-\sin^2(\theta)}$, we have $\cos(\theta) = -\sqrt{1-k^2}$ because the cosine of an obtuse angle is negative. Thus, $$\tan(\theta+90^{\circ}) = -\frac{-\sqrt{1-k^2}}{k} = \frac{\sqrt{1-k^2}}{k}.$$