How do you find the exact value of $tan [ arctan (1/6) + arccos (3/5) ]$ ?

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Answer 1 · 2016-06-21T14:30:27

#27/14 and -21/22.

Let $a = arc tan (1/6)$ . Then, $tan a = 1/6>0$ .

a is in either first quadrant or in the third.

Let $b = arc cos (3/5)$ . Then, $cos b = 3/5>0$ .

b is in either first quadrant or in the fourth..

Accordingly, $sin a = .+-4/5 and tan a = +-4/3$ .

Now, the given expression is

$tan ( a + b )= (tan a + tan b )/(1-tana tan b)$

$= ((1/6)+-(4/3))/(1-(+-)(1/6)(4/3))$

$=27/14 and - 21/22$ .

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How do you find the exact value of tan [ arctan (1/6) + arccos (3/5) ]tan [ arctan (1/6) + arccos (3/5) ]?