Question

How do you evaluate `\tan ( \sin ^ { - 1} ( \frac { 4} { 5} ) - \cos ^ { - 1} ( \frac { 5} { 13} ) )`?

Answer 1

`tan(sin^-1(4/5)-cos^-1(5/13))=-16/63`

Explanation:

Note :

`color(blue)((A)tan^-1x-tan^-1y=tan^-1((x-y)/(1+xy))`

`color(brown)((B)tan^-1(-x)=-tan^-1x`

We know that,

`(1)sin^-1x=tan^-1(x/sqrt(1-x^2))=>`

`sin^-1(4/5)=tan^-1((4/5)/sqrt(1-16/25))=tan^-1((4/5)/(3/5))`

`color(red)(sin^-1(4/5)=tan^-1(4/3)`

`(2)cos^-1x=tan^-1(sqrt(1-x^2)/x)=>`

#cos^-1(5/13)=tan^-1(sqrt(1- 25/169)/(5/13))=tan^-1((12/13)/(5/13))#

`color(red)(cos^-1(5/13)=tan^-1(12/5)`

Now, from `(1) and (2)` ,above

`sin^-1(4/5)-cos^-1(5/13)=color(blue)(tan^-1(4/3)-tan^-1(12/5)`

#color(white)(sin^-1(4/5)-cos^-1(5/13))=color(blue)(tan^-1((4/3- 12/5)/(1+4/3*12/5))touse(A)#

#color(white)(sin^-1(4/5)-cos^-1(5/13))=tan^-1((20- 36)/(15+48))#

`color(white)(sin^-1(4/5)-cos^-1(5/13))=tan^-1(-16/63)`

`sin^-1(4/5)-cos^-1(5/13)=color(brown)(-tan^-1(16/63)toApply(B)`

So,

`tan(sin^-1(4/5)-cos^-1(5/13))=tan(-tan^-1(16/63))`

`tan(sin^-1(4/5)-cos^-1(5/13))=-tan(tan^-1(16/63)`

`tan(sin^-1(4/5)-cos^-1(5/13))=-16/63`