What does #cos(arctan(3))+sin(arctan(4))# equal?

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Answer 1 · 2018-03-04T04:02:18

#cos(arctan(3))+sin(arctan(4))=1/sqrt(10)+4/sqrt(17)#

Let #tan^-1(3)=x#

then #rarrtanx=3#

#rarrsecx=sqrt(1+tan^2x)=sqrt(1+3^2)=sqrt(10)#

#rarrcosx=1/sqrt(10)#

#rarrx=cos^(-1)(1/sqrt(10))=tan^(-1)(3)#

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

then #rarrtany=4#

#rarrcoty=1/4#

#rarrcscy=sqrt(1+cot^2y)=sqrt(1+(1/4)^2)=sqrt(17)/4#

#rarrsiny=4/sqrt(17)#

#rarry=sin^(-1)(4/sqrt(17))=tan^(-1)4#

Now, #rarrcos(tan^(-1)(3))+sin(tan^(-1)tan(4))#

#rarrcos(cos^-1(1/sqrt(10)))+sin(sin^(-1)(4/sqrt(17)))=1/sqrt(10)+4/sqrt(17)#