Target Mathematics

2.20: Tan–sec sum identity and sine equation

0606/22/O/N/19 — Question 6 · 9 marks

(i)
Show that tanx1+secx+1+secxtanx=2sinx\dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} = \dfrac{2}{\sin x}.
[5]
(ii)
Hence solve the equation tanx1+secx+1+secxtanx=1+3sinx\dfrac{\tan x}{1 + \sec x} + \dfrac{1 + \sec x}{\tan x} = 1 + 3\sin x for 0°x180°0^{\degree} \le x \le 180^{\degree}.
[4]