Question :
(i) \(2 tan 30\over 1 + tan^2 30\) =
(a) sin 60
(b) cos 60
(c) tan 60
(d) sin 30
(ii) \(1 – tan^2 45\over 1 + tan^2 45\) =
(a) tan 90
(b) 1
(c) sin 45
(d) 0
(iii) sin 2A = 2 sin A is the true when A =
(a) 0
(b) 30
(c) 45
(d) 60
(iv) \(2 tan 30\over 1 – tan^2 30\) =
(a) cos 60
(b) sin 60
(c) tan 60
(d) sin 30
Solution :
(i) (a) Because \(2 tan A\over 1 + tan^2 A\) = sin 2A = sin 60
(ii) (d) Because \(1 – tan^2 45\over 1 + tan^2 45\) = \(1 – 1\over 1 + 1\) = 0
(iii) (a) Because when A = 0, sin 2A = sin 0 = 0
and, 2 sin A = 2 sin 0 = 2(0) = 0
or sin 2A = 2 sin A, when A = 0
(iv) (c) Because \(2 tan A\over 1 – tan^2 A\) = tan 2A = tan 60
