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If (a)/(b)=tan x>1 and -(π)/(2)<x<(π)/(2), then tan⁻¹ ((acos x-bsin x)/(bcos x+asin x)) is equal to:

1. tan⁻¹(a/b - x)
2. tan⁻¹(a/b)-x
3. tan⁻¹(a/b)-1/x
4. tan⁻¹(a/b)-x

Correct answer: tan⁻¹(a/b)-x

Solution

Dividing through gives (a/b - tan x)/(1 + (a/b)tan x) = tan(tan^-1(a/b) - x), so the expression equals tan^-1(a/b) - x. (Note the correct value appears at both index 1 and index 3; stored index 0 'tan^-1(a/b - x)' is wrong.)

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