Powerful Trigonometric ratios!

Relevant wiki: Proving Trigonometric Identities - Intermediate

Let $a = \sin \theta$ and $b = \cos \theta$ . $\implies T_1 = a+b$ and $T_2 = a^2+b^2 = 1$ .

$\begin{aligned} T_3 & = (a+b)(a^2+b^2 - ab) = (a+b)(1-ab) \\ \implies T_6 & = T_3^2 - 2a^3b^3 = (a+b)^2(1-ab)^2 - 2a^3b^3 = 1-3a^2b^2 \\ T_4 & = T_2^2 - 2a^2b^2 = 1 - 2a^2b^2 \\ \implies T_8 & = T_4^2 - 2a^4b^4 = (1 - 2a^2b^2)^2 - 2a^4b^4 = 1-4a^2b^2+2a^4b^4 \\ \implies T_{10} & = T_4T_6 - a^4b^6-a^6b^4 = (1 - 2a^2b^2)(1 - 3a^2b^2) - a^4b^4(a^2+b^2) = 1-5a^2b^2 +5a^4b^4 \end{aligned}$

Therefore,

$\begin{aligned} 6T_{10}-15T_8+10T_6 & = 6\left(1-5a^2b^2 +5a^4b^4\right) - 15\left(1-4a^2b^2+2a^4b^4\right) + 10\left(1-3a^2b^2\right) \\ & = 6-15+10 +(-30+60-30)a^2b^2 +(30-30)a^4b^4 \\ & = \boxed{1} \end{aligned}$