How do you calculate #sin(cos^-1(5/13)+tan^-1(3/4))#?

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Answer 1 · 2018-05-18T05:43:42

#sin(cos^(-1)(5/13)+ tan^(-1)(3/4))=63/65#

Let #cos^(-1)(5/13)=x# then

#rarrcosx=5/13#

#rarrsinx=sqrt(1-cos^2x)=sqrt(1-(5/13)^2)=12/13#

#rarrx=sin^(-1)(12/13)=cos^(-1)(5/13)#

Also, let #tan^(-1)(3/4)=y# then

#rarrtany=3/4#

#rarrsiny=1/cscy=1/sqrt(1+cot^2y)=1/sqrt(1+(4/3)^2)=3/5#

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

#rarrcos^(-1)(5/13)+ tan^(-1)(3/4)#

#=sin^(-1)(12/13)+sin^(-1)(3/5)#

#=sin^(-1)(12/13*sqrt(1-(3/5)^2)+3/5*sqrt(1-(12/13)^2))#

#=sin^(-1)(12/13*4/5+3/5*5/13)=63/65#

Now, #sin(cos^(-1)(5/13)+ tan^(-1)(3/4))#

#=sin(sin^(-1)(63/65))=63/65#