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

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Answer 1 · 2018-06-04T19:27:34

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

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#