Trigo dilemma

Apply $\csc(A) = \sec(90^\circ - A)$ and $\tan(A) = \cot(90^\circ - A)$ followed by $\cot(A) = -\cot(180^\circ - A)$ , and $\csc^2(A) = \cot^2(A) + 1$ .

$\begin{aligned} &&\left(\sec^{2}10^{\circ}+\tan 10^{\circ}\right)\left(\sec^{2} 50^{\circ} - \tan 50^{\circ}\right)\left(\sec^{2} 70^{\circ}+\tan 70^{\circ}\right) \\ &=& \left(\csc^{2}80^{\circ}+\cot 80^{\circ}\right)\left(\csc^{2} 40^{\circ} - \cot 40^{\circ}\right)\left(\csc^{2} 20^{\circ}+\cot 20^{\circ}\right) \\ &=& \left(\csc^{2}80^{\circ}+\cot 80^{\circ}\right)\left(\csc^{2} 140^{\circ} + \cot 140^{\circ}\right)\left(\csc^{2} 20^{\circ}+\cot 20^{\circ}\right) \\ &=& \left(\cot^{2}80^{\circ}+\cot 80^{\circ}+1\right)\left(\cot^{2} 140^{\circ} + \cot 140^{\circ}+1 \right)\left(\cot^{2} 20^{\circ}+\cot 20^{\circ} + 1\right) \\ \end{aligned}$

Note that

$\begin{aligned} \cot^2(A) + \cot(A) + 1 & =& \frac{\sin^2(A)}{\sin^2(A)} \left ( \left( \frac{\cos^2 A}{\sin^2 A} \right ) + \frac{\cos A}{\sin A} + 1 \right) \\ &=& \frac1{\sin^2(A)} ( \cos^2(A) + \sin(A) \cos(A) + \sin^2 (A) ) \\ &=& \frac1{\sin^2(A)} \left( 1 +\frac12 \sin(2A) \right ) \\ &=& \frac1{2\sin^2(A)} \left( \sin(2A) + 2\right ) \\ \end{aligned}$

Then,

$\begin{aligned} && \left(\sec^{2}10^{\circ}+\tan 10^{\circ}\right)\left(\sec^{2} 50^{\circ} - \tan 50^{\circ}\right)\left(\sec^{2} 70^{\circ}+\tan 70^{\circ}\right) \\ &=& \left(\cot^{2}80^{\circ}+\cot 80^{\circ}+1\right)\left(\cot^{2} 140^{\circ} + \cot 140^{\circ}+1 \right)\left(\cot^{2} 40^{\circ}+\tan 40^{\circ} + 1\right) \\ &=& \frac18 \frac{\bigg [ ((\sin(280^\circ) + 2)(\sin(160^\circ) + 2)(\sin(40^\circ) + 2) \bigg ] }{(\sin(20^\circ) \sin(140^\circ) \sin(80^\circ) )^2}\\ &=& \frac18 \frac{\bigg [ ((\sin(280^\circ) + 2)(\sin(160^\circ) + 2)(\sin(40^\circ) + 2) \bigg ] }{(\sin(20^\circ) \sin(140^\circ) \sin(260^\circ) )^2} \qquad (\ast) \\ \end{aligned}$

It's easier to evaluate the denominator of $\ast$ first. We see that $20 \times 3 = 60, 140 \times 3 = 360 + 60, 260 \times 3 = 2\times 360 + 60$ . Consider $\sin(3x) = \frac{\sqrt 3}2$ then angles $20^\circ, 140^\circ,260^\circ$ satisfy the equation. By Triple Angle formula: $-4\sin^3(x) + 3\sin(x) - \frac{\sqrt 3}2 = 0$ , and by Vieta's formula, $\sin(20^\circ) \sin(140^\circ) \sin(260^\circ) = \frac{\sqrt 3}8$ .

Similarly, for the numerator, we see that $280\times3= 2\times360 +120, 160\times3=360+120, 40\times3 =120$ . Consider $\sin(3y) = \frac{\sqrt 3}2$ then angles $40^\circ, 160^\circ,280^\circ$ satisfy the equation. By Triple Angle formula again: $-4\sin^3(y) + 3\sin(y) - \frac{\sqrt 3}2 =0$ . Letting $z = \sin(y)$ , then $8z^3 - 6z - \sqrt3 = 0$ have roots $\sin(40^\circ), \sin(160^\circ), \sin(280^\circ)$ . Equivalently, $8(z-2)^3 - 6(z-2) - \sqrt3 = 0$ have roots $\sin(40^\circ) + 2, \sin(160^\circ)+2, \sin(280^\circ) +2$ . Thus by Vieta's formula, we have $((\sin(280^\circ) + 2)(\sin(160^\circ) + 2)(\sin(40^\circ) + 2)$ equals to the negative ratio of constant term to the leading coefficient of the previous equation, which is $\frac18(-2^3 \cdot 8 + 6\cdot 2 - \sqrt3) = \frac18 (52- \sqrt 3)$ .

Putting it all together gives

$\begin{aligned} &&\left(\sec^{2}10^{\circ}+\tan 10^{\circ}\right)\left(\sec^{2} 50^{\circ} - \tan 50^{\circ}\right)\left(\sec^{2} 70^{\circ}+\tan 70^{\circ}\right) \\ &=& \frac18 \cdot \frac18 (52- \sqrt 3) \div \left( \frac {\sqrt 3}2 \right)^2 = \large \boxed{\frac{52-\sqrt3}3} \\ \end{aligned}$

THESE ARE FEW HINTS TO SOLVE THIS PROBLEM IN LESS STEPS First convert all terms into sin and cos and solve numerator and denominator separately .

For numerator: convert all trig ratios in acute angle of sine by using quadrants and identity 2sinxcosx=sin(2x). For simple calculations multiply 2 factors having positive signs and apply C and D formulaes. After simplifying multiply 3rd term . You will get numerical value except -sin20 sin40 sin80. You can solve it seperately my using sinx sin(60+ x) sin(60-x)=0.25 sin(3x). For denominator : use cosx cos(60+ x) cos(60-x)=0.25 cos(3x)

I know its a bit tedious but have a look please.

$\left(\sec^{2}10^{\circ}+\tan 10^{\circ}\right)\left(\sec^{2} 50^{\circ} - \tan 50^{\circ}\right)\left(\sec^{2} 70^{\circ}+\tan 70^{\circ}\right)$ $\begin{aligned}\\&=\left(\frac{1+\sin10^{\circ}\cos10^{\circ}}{\cos^{2} 10^{\circ}}\right)\left(\frac{1-\sin50^{\circ}\cos50^{\circ}}{\cos^{2} 50^{\circ}}\right)\left(\frac{1+\sin70^{\circ}\cos70^{\circ}}{\cos^{2}70^{\circ}}\right)\\&=\left(\frac{2+\sin20^{\circ}}{2\cos^{2} 10^{\circ}}\right)\left(\frac{2-\sin100^{\circ}}{2\cos^{2}50^{\circ}}\right)\left(\frac{2+\sin140^{\circ}}{2\cos^{2}70^{\circ}}\right)=\frac{x}{y}\end{aligned}$ $\\\\$

Now, we evaluate $x$ $\\\\$

$\begin{aligned}x&=\left(2+\sin20^{\circ}\right)\left(2-\sin100^{\circ}\right)\left(2+\sin140^{\circ}\right)\\&=\left(4-2\left(\sin100^{\circ}-\sin20^{\circ}\right)-\sin100^{\circ}\sin20^{\circ}\right)\left(2+\sin140^{\circ}\right)\\&=\left(4-4\cos60^{\circ}\sin40^{\circ}+\frac{1}{2}\left(\cos120^{\circ}-\cos80^{\circ}\right)\right)\left(2+\sin140^{\circ}\right)\\&=\left(\frac{15}{4}-2\sin40^{\circ}-\frac{1}{2}\cos80^{\circ}\right)\left(2+\sin140^{\circ}\right)\\&=\frac{15}{2}+\frac{15}{4}\sin140^{\circ}-4\sin40^{\circ}-2\sin140^{\circ}\sin40^{\circ}-\cos80^{\circ}-\frac{1}{2}\sin140^{\circ}\cos80^{\circ}\\&=\frac{15}{2}-\frac{1}{4}\sin40^{\circ}-2\sin^{2}40^{\circ}-\left(1-2\sin^{2}40^{\circ}\right)-\frac{1}{4}\left(\sin220^{\circ}+\sin60^{\circ}\right)\\&=\frac{13}{2}-\frac{1}{4}\sin40^{\circ}+\frac{1}{4}\sin40^{\circ}-\frac{\sqrt{3}}{8}=\frac{52-\sqrt{3}}{8}\end{aligned}$

Now, we evaluate $y$

$\begin{aligned}2\cos^{2}10^{\circ}\times2\cos^{2}50^{\circ}\times2\cos^{2}70^{\circ}&=8\left(\cos10^{\circ}\times\cos50^{\circ}\times\cos70^{\circ}\right)^{2}\\&=8\left(\frac{(\cos60^{\circ}+\cos40^{\circ})}{2}\times\cos70^{\circ}\right)^{2}\\&=8\left(\frac{1}{4}\cos70^{\circ}+\frac{1}{4}(\cos110^{\circ}+\cos30^{\circ})\right)^{2}\\&=8\left(\frac{\sqrt{3}}{8}+\frac{1}{4}(\cos110^{\circ}+\cos70^{\circ})\right)^{2}=8\times\frac{3}{64}=\frac{3}{8}\end{aligned}$

Now, we have

$\large{\frac{x}{y}=\frac{52-\sqrt{3}}{3}=\frac{a-\sqrt{b}}{b}}$

Finally, $\boxed{a+b^2=61}$

$LaTeX$ \begin{align } &\left(\sec^2 10°+\tan 10°\right)\left(\sec^2 50°-\tan 50°\right)\left(\sec^2 70°+\tan 70°\right)\ =&\left(\sec^2 10°+\tan 10°\right)\left(\sec^2 130°+\tan 130°\right)\left(\sec^2 70°+\tan 70°\right) \end{align }

For all $\theta \neq 90 + k\times 180, k\in Z$, $\sec^2\theta = 1 + \tan^2\theta$ so,\ \begin{align } =&\left(1+\tan 10°+\tan^2 10°\right)\left(1+\tan 130°+\tan ^2130°\right)\left(1+\tan 70°+\tan ^270°\right) \end{align } we note $\tan 10° = a, \tan 130° = b, \tan 70° = c$\

we have to calculate $(1+a+a^2)(1+b+b^2)(1+c+c^2)$\ we develop (tedious calculation) : \ $1 + (a+b+c)+(ab+ac+bc)+(ab^2+ac^2+a^2b+a^2c+b^2c+bc^2)+(a^2+b^2+c^2)\+(a^2b^2+a^2c^2+b^2c^2)+(abc)+(ab^2c+a^2bc+abc^2)+(a^2b^2c+a^2bc^2+ab^2c^2)+(a^2b^2c^2)$ (1)\ with Samarth Kapoor's idea : $(x-a)(x-b)(x-c) = x^3-\sqrt{3}x^2-3x+\dfrac{1}{\sqrt{3}}$\ we get \begin{align } a+b+c&=\sqrt{3}\ ab+ac+bc&=-3\ abc&=-\dfrac{1}{\sqrt{3}} \end{align } so we find \begin{align } a^2+b^2=c^2&=9\ ab^2+ac^2+a^2b+a^2c+bc^2+b^2c&=-2\sqrt{3}\ a^2b^2+a^2c^2+b^2c^2&=11\ ab^2c+a^2bc+abc^2&=-1\ a^2b^2c+a^2bc^2+ab^2c^2&=\sqrt{3}\ a^2b^2c^2&=\dfrac{1}{3} \end{align } Thus, we report in (1) and we get $(1+a+a^2)(1+b+b^2)(1+c+c^2)=\dfrac{52-\sqrt{3}}{3}$