Question

=
${ \sin}^{2} 12 - \frac{1}{ { \tan }^{2} 78?}$
​

Answer 1

Answer:

${sin}^{2}12 - \frac{1}{{tan}^{2}78} \\ \\ \sf\implies\: {sin}^{2} 12 - \frac{1}{{tan}^{2}(90 - 12)} \\ \\\sf\implies\: {sin}^{2}12 - \frac{1}{{cot}^{2}12} \\ \\\sf\implies\: {sin}^{2}12 - \frac{1}{\frac{{cos}^{2}12}{{sin}^{2}12}}$

Answer 2

Hey!

${sin}^{2} 12</strong><strong>°</strong><strong> </strong><strong>\: - \frac{1}{ {tan}^{2} 78</strong><strong>°</strong><strong>}$

${sin}^{2} 12</strong><strong>°</strong><strong> - \frac{ {cos}^{2}78</strong><strong>°</strong><strong> }{ {sin}^{2}78</strong><strong>°</strong><strong>}$

$\frac{{sin}^{2}12</strong><strong>°</strong><strong> \times { {sin}^{2} }78</strong><strong>°</strong><strong>- {cos}^{2} 78</strong><strong>°</strong><strong>}{ {sin}^{2}78</strong><strong>°</strong><strong>}$

$\frac{ {sin}^{2}12</strong><strong>°</strong><strong> \times {cos}^{2}12</strong><strong>°</strong><strong> - {cos}^{2}78</strong><strong>°</strong><strong>}{ {sin}^{2}78</strong><strong>°</strong><strong>}$

$\frac{ {sin}^{2}6</strong><strong>°</strong><strong> - {cos}^{2}78</strong><strong>°</strong><strong> }{ {sin}^{2}78</strong><strong>°</strong><strong>}$

OR

• sin²12° - 1/tan²78°

• sin²12° - 1/cot²12°

• sin²12° - 1/cos²12°/sin²12°

• sin²12° - sin²12°/cos²12°

• sin²12°/cos²12° - sin²12°/cos²12°

• tan²12° - tan²12°

• 0

Hope it helps!