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Determinant

Prove the fol...

Question

Prove the following identities
$2 (s i n^{6} θ + c o s^{6} θ) - 3 (s i n^{4} θ + c o s^{4} θ) + 1 = 0$

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Solution

$2 (s i n ⁶ θ + c o s ⁶ θ) - 3 (s i n ⁴ θ + c o s ⁴ θ) + 1 = 0$

$L H S = 2 (s i n ⁶ θ + c o s ⁶ θ) - 3 (s i n ⁴ θ + c o s ⁴ θ) + 1$

$= 2 [(s i n ² θ) ³ + (c o s ² θ) ³] - 3 s i n ⁴ θ - 3 c o s ⁴ θ + 1$

$= 2 (s i n ² θ + c o s ² θ) [(s i n ² θ) ² + (c o s ² θ) ² - s i n ² θ c o s ² θ)] - 3 s i n ⁴ θ - 3 c o s ⁴ θ + 1$ .....As $[a ³ + b ³ = (a + b) (a ² + b ² - a - b]$

$= 2 (1) [s i n ⁴ θ + c o s ⁴ θ - s i n ² θ c o s ² θ] - 3 s i n ⁴ θ - 3 c o s ⁴ θ + 1$ ...........As $[s i n ² θ + c o s ² θ = 1]$

$= 2 s i n ⁴ θ + 2 c o s ⁴ θ - 2 s i n ² θ c o s ² θ - 3 s i n ⁴ θ - 3 c o s ⁴ θ + 1$

$= 2 s i n ⁴ θ - 3 s i n ⁴ θ + 2 c o s ⁴ θ - 3 c o s ⁴ θ - 2 s i n ² θ c o s ² θ + 1$

$= - s i n ⁴ θ - c o s ⁴ θ - 2 s i n ² θ c o s ² θ + 1$

$= 1 - [s i n ⁴ θ + c o s ⁴ θ + 2 s i n ² θ (c o s ² θ)]$

$= 1 - [(s i n ² θ) ² + (c o s ² θ) ² + 2 s i n ² θ c o s ² θ]$

$= 1 - (s i n ² θ + c o s ² θ) ²$ ..............As $[a ² + b ² + 2 a b = (a + b) ²]$
$= 1 - (1) ²$

$L H S = 1 - 1 = 0$

$L H S = R H S$

$2 (s i n ⁶ θ + c o s ⁶ θ) - 3 (s i n ⁴ θ + c o s ⁴ θ) + 1 = 0$

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Similar questions

\begin{matrix} 2 ({sin}^{6} θ + {cos}^{6} θ) & - 3 ({sin}^{4} θ + {cos}^{4} θ) + 1 = 0 \end{matrix}

Use the suitable identity and simplify the given expression. $2 (s i n^{6} θ + c o s^{6} θ) - 3 (s i n^{4} θ + c o s^{4} θ) + 1$

2 (s i n^{6} θ + c o s^{6} θ) - 3 (s i n^{4} θ + c o s^{4} θ)

is equal to _____.

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

Q. 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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