Question

sin^4θ/a + cos^4θ/6 = 1/a+6
then find
sin^12θ/a^3 + cos^12θ/6^3​

Answer 1

Answer:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ \frac{{ \sin}^{4} \theta}{a} + \frac{{ \cos}^{4} \theta}{6} = \frac{1}{a + 6} \\ \\ \blue{\mathfrak{\underline{\large{To \: find}}}:} \\ \frac{{ \sin}^{12} \theta}{ {a}^{3} } + \frac{{ \cos}^{12} \theta}{ {6}^{3} } \\ \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ \frac{{ \sin}^{4} \theta}{a} + \frac{{ \cos}^{4} \theta}{6} = \frac{1}{a + 6} \\ \blue{\mathfrak{\underline{\large{cube \: on \: both \:sides }}}:} \\ {( \frac{{ \sin}^{4} \theta}{a} + \frac{{ \cos}^{4} \theta}{6})}^{3} = {( \frac{1}{a + 6} )}^{3} \\ \\ \frac{{ \sin}^{12} \theta}{ {a}^{3} } + \frac{{ \cos}^{12} \theta}{ {6}^{3} } + 3 \frac{{ \sin}^{8} \theta}{ {a}^{2} } \times \frac{{ \cos}^{4} \theta}{ 6 } + 3 \frac{{ \sin}^{4} \theta}{ a } \times \frac{{ \cos}^{8} \theta}{ {6}^{2} } = \frac{1}{ {(a + 6)}^{3} } \\ \\ \frac{{ \sin}^{12} \theta}{ {a}^{3} } + \frac{{ \cos}^{12} \theta}{ {6}^{3} } + 3\frac{{ \sin}^{4} \theta}{a} \times \frac{{ \cos}^{4} \theta}{6}(\frac{{ \sin}^{4} \theta}{a} + \frac{{ \cos}^{4} \theta}{6}) = \frac{1}{ {(a + 6)}^{3} } \\ \\ \frac{{ \sin}^{12} \theta}{ {a}^{3} } + \frac{{ \cos}^{12} \theta}{ {6}^{3} } + 3\frac{{ \sin}^{4} \theta}{a} \times \frac{{ \cos}^{4} \theta}{6}( \frac{1}{a + 6} ) =\frac{1}{ {(a + 6)}^{3} } \\ \\ \frac{{ \sin}^{12} \theta}{ {a}^{3} } + \frac{{ \cos}^{12} \theta}{ {6}^{3} } = \frac{1}{ {(a + 6)}^{3} } - \frac{{ \sin}^{4} \theta{ \cos}^{4} \theta}{2a(a + 6)} \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$