Find the value

Let $t = \tan{75^\circ}$ , then we have $\cot{75^\circ} = \dfrac{1}{t}$ and:

$\begin{aligned} \frac{\cot^3{75^\circ} + \tan^3{75^\circ}} {\cot{75^\circ} + \tan{75^\circ}} & = \frac{t^3 + \frac{1}{t^3}} {t + \frac{1}{t}} \\ & = \frac{\left(t + \frac{1}{t} \right) \left(t^2 -1 + \frac{1}{t^2}\right)} {t + \frac{1}{t}} \\ & = t^2 -1 + \frac{1}{t^2} \\ & = \left(t +\frac{1}{t}\right)^2 - 3 \\ & = \left(\tan{75^\circ} + \frac{1}{\tan {75 ^ \circ }}\right)^2 - 3 \\ & = \left( \frac{ \sin^2{75^\circ} + \cos^2{75^\circ}} {\sin{75^\circ} \cos{75^\circ}}\right)^2 - 3 \\ & = \left(\frac{2} {\sin{150^\circ}}\right)^2 - 3 \\ & = \left(\frac{2} {\sin{30^\circ}}\right)^2 - 3 \\ & = \left(4\right)^2 - 3 \\ & = 16 - 3 = \boxed{13} \end{aligned}$

Using the fact that $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}),$ where in this case $a = \cot(75^{\circ})$ and $b = \tan(75^{\circ}),$ we see that the given expression can be simplified to

$\cot^{2}(75^{\circ}) - \cot(75^{\circ})\tan(75^{\circ}) + \tan^{2}(75^{\circ}) =$

$(\csc^{2}(75^{\circ}) - 1) - 1 + (\sec^{2}(75^{\circ}) - 1) =$

$\dfrac{1}{\sin^{2}(75^{\circ})} + \dfrac{1}{\cos^{2}(75^{\circ})} - 3 =$

$\dfrac{\cos^{2}(75^{\circ}) + \sin^{2}(75^{\circ})}{\sin^{2}(75^{\circ})\cos^{2}(75^{\circ})} - 3 =$

$\dfrac{4}{(2\sin(75^{\circ})\cos(75^{\circ}))^{2}} - 3 =$

$\dfrac{4}{\sin^{2}(150^{\circ})} - 3 = \dfrac{4}{(\frac{1}{2})^{2}} - 3 = 4*4 - 3 = \boxed{13}.$

Observe the expression is factorable.

$\Large \frac { a^3 +b^3 }{ a+b} \normalsize = \Large \frac{(a+b) * (a^2 - ab + b^2)}{(a+b)}$ $................. [ Taking$ $cot 75^\circ = a , tan 75^\circ = b ]$

$= a^2 + b^2 - 1$ $.................................. [ Since$ $ab = cot 75^\circ * tan 75^\circ = 1 ]$

$= cot^2 75^\circ + tan^2 75^\circ - 1$

$= cot^2 (90^\circ - 15^\circ ) + tan^2 75^\circ - 1$

$= tan^2 15^\circ + tan^2 75^\circ - 1$

$= \Large \frac{sin^2 15^\circ}{cos^2 15^\circ} \normalsize + \Large \frac{sin^2 75^\circ}{cos^2 75^\circ} \normalsize - 1$

$= \Large \frac{sin^2 15^\circ}{cos^2 15^\circ} \normalsize - 1 + 1 + \Large \frac{sin^2 (90^\circ - 15^\circ )}{cos^2 (90^\circ - 15^\circ )} \normalsize - 1$

$= 1 - \{ 1 - \Large \frac{sin^2 15^\circ}{cos^2 15^\circ} \normalsize \} + \{ \Large \frac{cos^2 15^\circ}{sin^2 15^\circ} \normalsize - 1 \}$

$= 1 - \{ \Large \frac{cos^2 15^\circ - sin^2 15^\circ}{cos^2 15^\circ} \normalsize \} + \{ \Large \frac{cos^2 15^\circ - sin^2 15^\circ }{sin^2 15^\circ} \normalsize \}$

$= 1 - \{ \Large \frac{cos (2 *15^\circ )}{cos^2 15^\circ} \normalsize \} + \{ \Large \frac{cos (2 *15^\circ )}{sin^2 15^\circ} \normalsize \}$

$= 1 - \Large \frac{cos 30^\circ }{cos^2 15^\circ} \normalsize + \Large \frac{cos 30^\circ }{sin^2 15^\circ}$

$= 1 + cos 30^\circ \Large ( \frac{1 }{sin^2 15^\circ} \normalsize - \Large \frac{1}{cos^2 15^\circ} )$

$= 1 + cos 30^\circ \Large ( \frac{cos^2 15^\circ - sin^2 15^\circ }{sin^2 15^\circ * cos^2 15^\circ} )$

$= 1 + cos 30^\circ * 4 * \Large \frac{cos (2 *15^\circ )}{4 * sin^2 15^\circ * cos^2 15^\circ}$

$= 1 + cos 30^\circ * 4 * \Large \frac{cos 30^\circ }{ sin^2 (2*15^\circ)}$

$= 1 + 4 * \Large \frac{cos^2 30^\circ }{ sin^2 30^\circ}$

$= 1 + 4 * \Large \frac{ ( \frac{\sqrt{3}}{2} ) ^2 }{ ( \frac{1}{2} ) ^2 }$

$= 1 + 4 * \Large \frac{ ( \frac{3}{4} ) }{ ( \frac{1}{4} ) }$

$= 1 + 4 * 3$

$= 13$

tan75=tan(30+45)=sin(30+45)÷cos(30+45). =((1÷2)×(1÷√2)+(√3÷2)×(1÷√2))÷ ((1÷√2)×(√3÷2)-(1÷√2)×(1÷2))=(√3+1)÷(√3-1). so cot75=(√3-1)÷(√3+1). (cot^3 (75)+tan^3(75))÷(cot75+tan75). =((√3-1)÷(√3+1))^3+((√3+1)÷(√3-1))^3÷ ((√3-1)÷(√3+1))+((√3+1)÷(√3-1)). Put x=√3-1÷√3+1 so 1÷x=√3+1÷√3-1. expression=x^3+(1÷x^3)÷((x)+(1÷x)). x+(1÷x)=((√3-1)÷(√3+1))+((√3+1)÷(√3-1)). =((√3+1)^2+(√3-1)^2)÷((√3-1)×(√3+1)). =(3+1+2√3+3+1-2√3)÷(3-1)=4. (x+(1÷x))^3=x^3+(1÷x)^3+3(x+(1÷x). x^3+(1÷x)^3=64-3×4=52. So. Expression=52÷4=13#######

irst factor the numerator as a sum of two cubes, and simplify. You'll get the following.

(cot^3(75) + tan^3(75))/(cot(75) + tan(75)) = cot^2(75) -cot(75)tan(75) + tan^2(75) = cot^2(75) -1 + tan^2(75)

Next use the identity tan^2(u) = (1 - cos(2u))/(1 + cos(2u)), where u = 75.

tan^2(75) = (1 - cos(150))/(1 + cos(150)) = (1 + sqrt(3)/2)/(1 - sqrt(3)/2)) = 7 - 4sqrt(3)

cot^2(75) = (tan^2(75))^(-1) = (7 - 4sqrt(3))^(-1) = 7 + 4sqrt(3)

So putting it all together, I get the following.

(cot^3(75) + tan^3(75))/(cot(75) + tan(75)) = 7 + 4sqrt(3) - 1 + 7 - 4 sqrt(3) = 13

tan75=tan(45+30)=(sqrt3+1)/(sqrt3-1)= 2+sqrt3. tan^2 75=7-sqrt3. cot^2 75= 7+sqrt3 LHS=cot^2 75-cot75tan75+ tan^2 75 =7+sqrt3 - 1 +7-sqrt3 =13