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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