Question

Help with this trigonometry sum pleeeaase

Answer 1

Answer:

$\sqrt{3}$

Step-by-step explanation:

$\frac{ \tan(10) \tan(20) \tan(60) \tan(70) \tan(80) }{ { \sec(52) }^{2} - \ { \cot(38) }^{2} } \\ = \frac{ \tan(90 - 80) \tan(90 - 70) (\sqrt{3} ) \tan(70) \tan(80) }{ { \sec(90 - 38) }^{2} - { \cot(38) }^{2} } \\ \\ = \frac{ \: \: \cot(80) \cot(70) \times \sqrt{3} \tan(70) \tan(80) \: \: }{ { \csc(38) }^{2} - { \cot(38) }^{2} } \\ \\ = \sqrt{3}$