If $T_{n}=\sin ^{n} x+\cos ^{n} x$, prove that
(i) $\frac{T_{3}-T_{5}}{T_{1}}=\frac{T_{5}-T_{7}}{T_{3}}$
(ii) $2 T_{6}-3 T_{4}+1=0$
(iii) $6 T_{10}-15 T_{8}+10 T_{6}-1=0$
(i) LHS:
$\frac{T_{3}-T_{5}}{T_{1}}=\frac{\left(\sin ^{3} x+\cos ^{3} x\right)-\left(\sin ^{5} x+\cos ^{5} x\right)}{\sin x+\cos x}$
$=\frac{\sin ^{3} x-\sin ^{5} x+\cos ^{3} x-\cos ^{5} x}{\sin x+\cos x}$
$=\frac{\sin ^{3} x\left(1-\sin ^{2} x\right)+\cos ^{3} x\left(1-\cos ^{2} x\right)}{\sin x+\cos x}$
$=\frac{\sin ^{3} x \cdot \cos ^{2} x+c \operatorname{os}^{3} x \cdot \sin ^{2} x}{\sin x+\cos x}$
$=\frac{\sin ^{2} x \cdot \cos ^{2} x(\sin x+c \cos x)}{\sin x+\cos x}$
$=\sin ^{2} x \cdot \cos ^{2} x$
RHS:
$\frac{T_{5}-T_{7}}{T_{3}}$
$=\frac{\left(\sin ^{5} x+\cos ^{5} x\right)-\left(\sin ^{7} x+\cos ^{7} x\right)}{\sin ^{3} x+\cos ^{3} x}$
$=\frac{\sin ^{5} x-\sin ^{7} x+\cos ^{5} x-\cos ^{7} x}{\sin ^{3} x+\cos ^{3} x}$
$=\frac{\sin ^{5} x\left(1-\sin ^{2} x\right)+\cos ^{5} x\left(1-\cos ^{2} x\right)}{\sin ^{3} x+\cos ^{3} x}$
$=\frac{\sin ^{5} x \cos ^{2} x+\cos ^{5} x \sin ^{2} x}{\sin ^{3} x+\cos ^{3} x}$
$=\sin ^{2} x \cdot \cos ^{2} x$
LHS = RHS
Hence proved.
(ii) LHS:
$2 T_{6}-3 T_{4}+1$
$2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
$2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
2. 1. $\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
$2 \sin ^{4} x+2 \cos ^{4} x-2 \sin ^{2} x \cos ^{2} x-3 \sin ^{4} x-3 \cos ^{4} x+1$
$-\left(\sin ^{4} x+\cos ^{4} x\right)-\sin ^{2} x \cos ^{2} x+1$
$-\left(\sin ^{2} x+\cos ^{2} x\right)^{2}+1$
$-1+1$
= 0
Hence proved.
(iii) LHS:
$6 T_{10}-15 T_{8}+10 T_{6}-1$
$6\left(\sin ^{10} x+\cos ^{10} x\right)-15\left(\sin ^{8} x+\cos ^{8} x\right)+10\left(\sin ^{6} x+\cos ^{6} x\right)-1$