How do you evaluate #cos(arctan(-2) + arccos(5/13))#?

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Answer 1 · 2016-07-30T08:23:48

#=29/(13sqrt 5)#

Let #a = arc tan (-2) in Q3#,

wherein sine and tangent are negative.

Then, #tan a = -2, sin a =-2/sqrt 5 and cos a = 1/sqrt 5#

Let #b = arc cos (5/13) in Q1#, wherein sine is positive..

Then, #cos b = 5/13 and sin b =12/13#.

Now, the given expression is

#cos(a+b)#

#=cos a cos b - sin a sin b#

#=(1/sqrt 5)(5/13)-(-2/sqrt 5)(12/13)#

#=29/(13sqrt 5)#