J's answer to Nancy's Secondary 3 A Maths Singapore question.
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Divide both numerator and denominator by cos x first.
(sec x - 2 cos x) / (sin x - cos x)
= ((sec x - 2 cos x)/cos x) / ((sin x - cos x)/cos x)
= (sec x (1/cos x) - 2) / (sin x / cos x - 1)
= (sec² x - 1 - 1) / (tan x - 1)
[Recall that tan x = sin x / cos x and 1/cos x = sec x]
= (tan² x - 1) / (tan x - 1)
[Recall that sec² x = tan² x + 1, so sec² x - 1 = tan² x]
= (tan x + 1)(tan x - 1) / (tan x - 1)
[Property a² - b² = (a + b)(a - b)]
= tan x + 1 (proved)
(sec x - 2 cos x) / (sin x - cos x)
= ((sec x - 2 cos x)/cos x) / ((sin x - cos x)/cos x)
= (sec x (1/cos x) - 2) / (sin x / cos x - 1)
= (sec² x - 1 - 1) / (tan x - 1)
[Recall that tan x = sin x / cos x and 1/cos x = sec x]
= (tan² x - 1) / (tan x - 1)
[Recall that sec² x = tan² x + 1, so sec² x - 1 = tan² x]
= (tan x + 1)(tan x - 1) / (tan x - 1)
[Property a² - b² = (a + b)(a - b)]
= tan x + 1 (proved)
Date Posted:
3 years ago
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