Given that tan 7 in quadrant IV nd sin 2 cos 2 and tan 2Sol

Given that tan = 7 in quadrant IV, nd sin 2, cos 2, and tan 2.

Solution

tan = 7

==> tan = 7/1

opposite side = 7 and adjacent side = 1

hypotenuse = [(-7)² + 1²] = 50

as it is in IV quadrant sine function is negative

==> hypotenuse = 52

sin 2 = 2 sin cos

==> sin 2 = 2 (-7/(52)) (1/ (52)) = -7/25

==> sin 2 = -7/25

cos 2 = cos² - sin²

==> cos 2 =(1/ (52))² - ( -7/(52))² = -6/50 = -3/25

==> cos 2 = -3/25

tan 2 = sin 2 / cos 2

==> tan 2 = (-7/25)/(-3/25)

==> tan 2 = 7/3

Hence sin 2 = -7/25 , cos 2 = -3/25 , tan 2 = 7/3