If $P_{n} = \cos^{n} θ + \sin^{n} θ, t h e n :$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2 P_{6} - 3 P_{4} = - 1$

$2 P_{6} + P_{1} = 1$

$6 P_{10} - 15 P_{8} + 10 P_{6} = 0$

$6 P_{10} - 15 P_{8} + 10 P_{6} = 1$

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P_{n} = \cos^{n} θ + \sin^{n} θ, t h e n :$

$\begin{array}{l} P_{n} = \cos^{n} θ + \sin^{n} θ \\ P_{n} - P_{n - 2} = \cos^{n} θ + \sin^{n} θ - \cos^{n - 2} θ - \sin^{n - 2} θ \\ P_{n} - P_{n - 2} = - \sin^{2} θ \cos^{2} θ P_{n - 4} \end{array}$

$\begin{aligned} 6 P_{10} - 15 P_{8} + 10 P_{6} - 1 = 6 (- \cos^{2} θ \sin^{2} θ P_{6} + P_{8}) - 15 (- \cos^{2} θ \sin^{2} θ P_{4} + P_{6}) + 10 P_{6} - 1 \\ = 6 (- \cos^{2} θ \sin^{2} θ (1 - 3 \sin^{2} θ \cos^{2} θ)) + (- \cos^{2} θ \sin^{2} θ (P_{4} + P_{6})) \\ - 15 (- \cos^{2} θ \sin^{2} θ (1 - 2 \sin^{2} θ \cos^{2} θ) + 1 - 3 \sin^{2} θ \cos^{2} θ) + 10 (1 - 3 \cos^{2} θ \sin^{2} θ) - 1 = o \end{aligned}$