Choose the correct answer from the alternatives given :
The value of the following is
$3 (s i n^{4} θ + c o s^{4} θ) + (s i n^{6} θ + c o s^{6} θ) + 12 s i n^{2} θ + c o s^{2} θ$

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Solution

The correct option is A 5
$3 (s i n^{4} θ + c o s^{4} θ) + (s i n^{6} θ + c o s^{6} θ) + 12 s i n^{2} θ + c o s^{2} θ$
$= 3 [(s i n^{2} θ + c o s^{2} θ)^{2} - 2 s i n^{2} θ c o s^{2} θ] + 2 [(s i n^{2} θ + c o s^{2} θ)^{3} - 3 (s i n^{2} θ c o s^{2} θ)] + 12 s i n^{2} θ c o s^{2} θ$
$= 3 (- 2 s i n^{2} θ c o s^{2} θ) + 2 (1 - 3 s i n^{2} θ c o s^{2} θ) + 12 s i n^{2} θ c o s^{2} θ$
= 5

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Similar questions

({sin}^{6} θ + {cos}^{6} θ) - 3 ({sin}^{4} θ + {cos}^{4} θ) + 1

c o s^{2} θ - s i n^{2} θ = \frac{1}{3}

0 \leq θ \leq \frac{π}{2}

c o s^{4} θ - s i n^{4} θ

2 ({sin}^{6} θ + {cos}^{6} θ) - 3 ({sin}^{4} θ + {cos}^{4} θ)

sin + {sin}^{2} θ = 1,

{cos}^{2} θ + {cos}^{4} θ

\sin^{6} θ + \cos^{6} θ = 1 - 3 \sin^{2} {θcos}^{2} θ

\sin^{2} θ + \cos^{4} θ = \cos^{2} θ + \sin^{4} θ

{cosec}^{4} θ - {cosec}^{2} θ = \cot^{4} θ + \cot^{2} θ

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