Question

What is the value of `tan(\cos ^ { - 1} \frac { 3} { 5} + \tan ^ { - 1} \frac { 1} { 4} )`?

Answer 1

`rarrtan^(-1)(cos^(-1)(3/5)+tan^(-1)(1/4))=19/8`

Explanation:

Let `cos^(-1)(3/5)=x` then

`rarrsecx=5/3`

`rarrtanx=sqrt(sec^2x-1)=sqrt((5/3)^2-1)=sqrt((5^2-3^2)/3^2)=4/3`

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

Now, using `tan^(-1)(A)+tan^(-1)(B)=tan^(-1)((A+B)/(1-AB))`

`rarrtan^(-1)(cos^(-1)(3/5)+tan^(-1)(1/4))`

`=tan^(-1)(tan^(-1)(4/3)+tan^(-1)(1/4))`

`=tan^(-1)(tan^(-1)((4/3+1/4)/(1-(4/3)*(1/4))))`

`=(19/12)/(8/12)=19/8`