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secondary 3 | A Maths
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pls help with this qn
Date Posted: 4 years ago
Views: 202
(1 + tanA)/(1 - tanA) + (1 - tanA)/(1 + tanA)
= [ (1 + tanA)(1 + tanA) + (1 - tanA)(1 - tanA)]/[(1 - tanA)(1 + tanA) ]
= [ (1 + tanA)² + (1 - tanA)² ] / (1² - tan²A)
= (1 + 2tanA + tan²A + 1 - 2tanA + tan²A)/)1 - tan²A)
= (2 + 2tan²A)/(1 - tan²A)
= (2 + 2sin²A/cos²A)/(1 - sin²A/cos²A)
= (2cos²A + 2sin²A)/(cos²A - sin²A)
(This step is like rationalising the denominator, multiplying the fraction by cos²A/cos²A)
= 2/cos2A
= 2sec2A
= [ (1 + tanA)(1 + tanA) + (1 - tanA)(1 - tanA)]/[(1 - tanA)(1 + tanA) ]
= [ (1 + tanA)² + (1 - tanA)² ] / (1² - tan²A)
= (1 + 2tanA + tan²A + 1 - 2tanA + tan²A)/)1 - tan²A)
= (2 + 2tan²A)/(1 - tan²A)
= (2 + 2sin²A/cos²A)/(1 - sin²A/cos²A)
= (2cos²A + 2sin²A)/(cos²A - sin²A)
(This step is like rationalising the denominator, multiplying the fraction by cos²A/cos²A)
= 2/cos2A
= 2sec2A
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