How do you find $tan (x+y)$ if $cot x = 6/5$ and $sec y = 3/2$ ?

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Answer 1 · 2015-05-15T05:43:37

To find tan (x + y), use the trig identity:
$tan (a + b) = (tan a + tan b)/(1 - tan a.tan b)$ (1)

First, find tan x and tan y.

$tan x = 1/cot x = 5/6$

$sec y = 1/cos y = 3/2 -> cos y = 2/3$

$sin^2 y = 1 - cos^2 y = 1 - 4/9 = 5/9 -> sin x = sqr5/3$

$tan y = sin y/cos y = ((sqr5)/3):(2/3) = (sqr5)/2$
Replace tan x and tan y into identity (1)

$f(x,y) = [(5/6 + (sqr5)/2]/[1 - 5sqr5/12)]$ =

$=(10 + 6sqr5)/(12 - 5sqr5)$

How do you find tan (x+y)tan (x+y) if cot x = 6/5cot x = 6/5 and sec y = 3/2sec y = 3/2?