Question

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

Answer 1

`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`