Please help me solve this plane trigonometry problem. plz?

Question

give the value of:

c. tan [ tan^-1 x+1/x-1 PLUS tan^-1 x-1/x ]

Answer 1

Remember that tan (x+y) = (tan x + tan y)/(1-tan x tan y), so:

tan (arctan ((x+1)/(x-1)) + arctan ((x-1)/x))
((x+1)/(x-1) + (x-1)/x)/(1-(x+1)/(x-1) * (x-1)/x)
((x+1)/(x-1) + (x-1)/x)/(1-(x+1)/x)
(x(x+1)/(x-1) + (x-1))/(x-(x+1))
-(x(x+1)/(x-1) + (x-1))
-(x(x+1) + (x-1)²)/(x-1)
-(x² + x + x² - 2x + 1)/(x-1)
(2x² - x + 1)/(1-x)

And we are done.

Answer 2

Use the tangent addition of angles formula.

tan(α + β) = (tanα + tanβ) / [1 - (tanα)(tanβ)]

tan { arctan[(x+1)/(x-1)] + arctan[(x-1)/x] }

= [(x + 1)/(x - 1) + (x - 1)/x] / {1 - [(x + 1)/(x - 1)]*[(x - 1)/x]}

Multiply numerator and denominator by x(x - 1).

= [x(x + 1) + (x - 1)²] / [x(x - 1) - (x + 1)(x - 1)]

= (x² + x + x² - 2x + 1) / (x² - x - x² + 1)

= (2x² - x + 1) / (-x + 1)

Answer 3

Where are you stuck? I won't do the homework for you, but I'll help explain something that's giving you trouble. tan^-1 is arctangent -- the angle that has "x" as its tangent.

Answer 4

2 - tanx