# Compute without calculator, $\frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} - \frac{1}{\cos^{2}(45)}$

Compute without calculator, $$\frac{1}{\cos^{2}(10^{\circ})} + \frac{1}{\sin^{2}(20^{\circ})} + \frac{1}{\sin^{2}(40^{\circ})} - \frac{1}{\cos^{2}(45^{\circ})}$$

Attempt:

Let $$A = \cos(10) \sin(20) \sin(40)$$, then

$$\frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} = \frac{\sin^{2}(20) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(20)}{A^{2}}$$ notice also $$2A \sin(10) \sin(40) = \sin^{2}(20) \sin^{2}(40)$$ $$2A \cos(20) \cos(10) = \cos^{2}(10) \sin^{2}(40)$$ so we have

$$\frac{2A \sin(10) \sin(40) + 2A \cos(20) \cos(10)+ \cos^{2}(10) \sin^{2}(20)}{A^{2}}$$

$$= \frac{2A \left[2 \sin(10) \sin(20) \cos(20) + \frac{\sqrt{3}}{2} + \sin(10)\sin(20) \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}}$$

$$= \frac{2A \left[\sin(10) \sin(20) (1+ \cos(20) + \cos(20)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}}$$

$$= \frac{2A \left[\sin(10) \sin(20) (3\cos^{2}(10) - \sin^{2}(10)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}}$$

How to continue then?

• Commented May 26, 2019 at 10:36

\begin{align*} &\;\frac{1}{\cos^{2}10^{\circ}} + \frac{1}{\sin^{2}20^{\circ}} + \frac{1}{\sin^{2}40^{\circ}} - \frac{1}{\cos^{2}45^{\circ}} \\ =&\;\frac {16\sin^210^\circ\cos^220^\circ}{16\sin^210^\circ\cos^210^\circ\cos^220^\circ}+\frac{4\cos^220^\circ}{4\sin^220^\circ\cos^220^\circ}+ \frac{1}{\sin^{2}40^{\circ}} -2\\ =&\;\frac{4(1-\cos20^\circ)(1+\cos40^\circ)+2(1+\cos40^\circ)+1}{\sin^240^\circ}-2\\ =&\;\frac{7-4\cos20^\circ+6\cos40^\circ-4\cos20^\circ\cos40^\circ}{\sin^240^\circ}-2\\ =&\;\frac{7-4\cos20^\circ+6\cos40^\circ-2\cos60^\circ-2\cos20^\circ}{\sin^240^\circ}-2\\ =&\;\frac{6-6\cos20^\circ+6\cos40^\circ}{\sin^240^\circ}-2\\ =&\;\frac{6-12\sin30^\circ\sin 10^\circ}{\sin^240^\circ}-2\\ =&\;\frac{6-6\cos 80^\circ}{\frac12(1-\cos80^\circ)}-2\\ =&\;10 \end{align*}

• wolframalpha.com/input/… Commented May 26, 2019 at 9:11
• @SoumalyaPramanik Your angles in the first three terms are incorrect. Commented May 26, 2019 at 9:13
• @SoumalyaPramanik You are wrong. Commented May 26, 2019 at 9:13
• $\cos A\cos B=\frac{1}{2}(\cos(A+B)+\cos(A-B))$. It is $-4\cos40^\circ\cos20^\circ=-2\cos60^\circ-2\cos20^\circ$. Commented May 26, 2019 at 9:40
• The double angle formula $\sin 2\theta=2\sin\theta\cos\theta$ and the facts that $40=2\times 20$ and $20=2\times 10$. These are big hints to me. Commented May 26, 2019 at 10:02

I believe this is how the problem came into being.

Observe that

$$\dfrac1{\cos^210^\circ}=1+\tan^210^\circ$$

$$\dfrac1{\sin^220^\circ}=\dfrac1{\cos^270^\circ}=1+\tan^270^\circ$$

$$\dfrac1{\sin^240^\circ}=\dfrac1{\cos^250^\circ}=1+\tan^2(-50^\circ)$$

Observe that the angles differ by $$60^\circ$$

Now if $$\tan3x=\tan3A,3x=180^\circ n+3A$$ where $$n$$ is any integer

$$x=60^\circ n+A$$ where $$n=-1,0,1$$ or more generally $$n\equiv-1,0,1\pmod3$$

Again, $$\tan3A=\tan3x=\dfrac{3t-t^3}{1-3t^2}$$

$$\implies t^3-(3\tan3A)t^2-3t+\tan3A=0$$ where $$t=\tan x$$

whose roots are $$t_r=\tan(60^\circ r+A)$$ where $$r=-1,0,1$$

$$\displaystyle\implies S(A)$$

$$\displaystyle=\sum_{r=-1}^1t_r^2 =\left(\sum_{r=-1}^1t_r\right)^2-2(t_{-1}t_0+t_0t_1+t_1t_{-1}) =\left(\dfrac{3\tan3A}1\right)^2-2\left(\dfrac{-3}1\right)=6+9\tan^23A$$

Here $$3A=30^\circ\equiv-150^\circ\equiv210^\circ\pmod{180^\circ}$$

So, $$\tan3A=\tan30^\circ=\dfrac1{\sqrt3}$$

$$\implies S(10^\circ)=6+9\left(\dfrac1{\sqrt3}\right)^2=?$$