If tan x + sec x = √3, 0 < x < π, then x is equal to
A. \(\cfrac{5\pi}6\)
B. \(\cfrac{2\pi}3\)
C. \(\cfrac{\pi}6\)
D.\(\cfrac{\pi}3\)
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If tan x + sec x = √3, 0 < x < π, then x is equal to
A. \(\cfrac{5\pi}6\)
B. \(\cfrac{2\pi}3\)
C. \(\cfrac{\pi}6\)
D.\(\cfrac{\pi}3\)
If tan x + sec x = √3, 0 < x < π, then x is equal to
A. \(\cfrac{5\pi}6\)
B. \(\cfrac{2\pi}3\)
C. \(\cfrac{\pi}6\)
D.\(\cfrac{\pi}3\)
Correct option is C. \(\cfrac{\pi}6\)
Given: tan x + sec x = √3
squaring on both sides
(tan x + sec x)2 = √32
tan2x+sec2x + 2 tan x sec x = 3
Also, sec2x – tan2x = 1
tan2x + 1+ tan2x+ 2tan x sec x = 3
2tan2x + 2tan x sec x = 3 – 1
tan2x + tan x sec x = 2/2
tan2x + tan x sec x = 1
tan x sec x = 1 – tan2x
again, squaring on both sides
tan2x sec2x = 1 + tan4x – 2 tan2 x
(1+ tan2x) tan2x = 1 + tan4x – 2 tan2x
Tan4x + tan2x = 1 + tan4x – 2 tan2x
3 tan2x = 1
tan x = 1/√3
x = π/6.