If Tn = sin^nθ + cos^nθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

Q: If T_n = sinⁿθ + cosⁿθ , then $\frac{T_6 – T_4}{T_6} = m $ holds for values of m satisfying

(A) $\large m \in [-1 , \frac{1}{3}]$

(B) $\large m \in [0 , \frac{1}{3}]$

(C) $\large m \in [-1 , 0]$

(D) $\large m \in [-1 , -\frac{1}{2}]$

Sol: $\large \frac{T_6 – T_4}{T_6} = \frac{-sin^2 2\theta}{4-3sin^2 2\theta} = m $

$\large sin^2 2\theta = \frac{4 m}{-1 + 3 m}$

$\large 0 \le \frac{4 m}{-1 + 3 m} \le 1 $

$\large m \in [-1 , 0]$

Hence (C), (D) are the correct answers