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

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Answer 1 · 2018-04-19T09:36:15

#rarrcos(sin^(-1)(4/5)+tan^(-1)(5/12))=16/65#

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#