How to find exact value COS(SIN^-1 4/5+TAN^-1 5/12) ?

1 Answer
Apr 19, 2018

rarrcos(sin^(-1)(4/5)+tan^(-1)(5/12))=16/65

Explanation:

Let sin^(-1)(4/5)=x then

rarrsinx=4/5

rarrtanx=1/cotx=1/(sqrt(csc^2x-1))=1/(sqrt((1/sinx)^2-1))=1/(sqrt((1/(4/5))^2-1))=4/3

rarrx=tan^(-1)(4/3)=sin^(-1)=(4/5)

Now,
rarrcos(sin^(-1)(4/5)+tan^(-1)(5/12))

=cos(tan^(-1)(4/3)+tan^(-1)(5/12))

=cos(tan^(-1)((4/3+5/12)/(1-(4/3)*(5/12))))

=cos(tan^(-1)((63/36)/(16/36)))

=cos(tan^(-1)(63/16))

Let tan^(-1)(63/16)=A then

rarrtanA=63/16

rarrcosA=1/secA=1/sqrt(1+tan^2A)=1/sqrt(1+(63/16)^2)=16/65

rarrA=cos^(-1)(16/65)=tan^(-1)(63/16)

rarrcos(sin^(-1)(4/5)+tan^(-1)(5/12))=cos(tan^(-1)(63/16))=cos(cos^(-1)(16/65))=16/65