Question

evaluate 3cos55°÷ 7sin35° - √3 ( tan 10° tan 30° tan 40° tan° 50 tan° 80)

Answer 1

$\frac{3 \cos(55) }{7 \sin(35) } - \sqrt{3} ( \tan(10) \tan(30) \tan(40) \tan(50) \tan(80) \\ \\ \frac{3 \cos(55) }{7 \sin(90 - 55) } - \sqrt{3} ( \tan(10) \tan(30) \tan(40) \tan(50) \tan(80) \\ \\ since \: \sin(90 - \alpha ) = \cos( \alpha ) \\ \\ therefore \\ \\ \frac{3 \cos(55) }{7 \cos(55) } - \sqrt{3} ( \tan(10) \tan(30) \tan(40) \tan(50) \tan(80) \\ \\ \frac{3}{7} - \sqrt{3} ( \tan(10) \tan(30) \tan(40) \tan(50) \tan(80) \\ \\ since \: \sqrt{3 } = \tan(60) \\ \\ \frac{3}{7} - \tan(60) \tan(10) \tan(30) \tan(40) \tan(50) \tan(80) \\ \\ \frac{3}{7 } \tan(90 - 30) \tan(90 - 80) \tan(30) \tan(90 - 50) \tan(50) \tan(80) \\ \\ since \: \tan(90 - \alpha ) = \cot( \alpha ) \\ \\ \\ \\ therefore \frac{3}{7 } - \cot(30) \cot(80) \tan(30) \cot(50) \tan(50) \tan(80) \\ \\ \frac{3}{7} - 1 \\ \\ = \frac{3 - 7}{7} \\ \\ = \frac{ - 4}{7}$