Find the value of #cos(tan^(-1)(2)+tan^(-1)(3))#?
2 Answers
Explanation:
First use the cosine angle-addition formula:
#cos(a+b)=cos(a)cos(b)-sin(a)sin(b)#
Then the original expression equals:
#=cos(tan^-1(2))cos(tan^-1(3))-sin(tan^-1(2))sin(tan^-1(3))#
Note that if
So, when
#cos(tan^-1(2))=cos(theta)="adjacent"/"hypotenuse"=1/sqrt5#
#sin(tan^-1(2))=sin(theta)="opposite"/"hypotenuse"=2/sqrt5#
Now let angle
#{("opposite"=3),("adjacent"=1),("hypotenuse"=sqrt10):}#
Then:
#cos(tan^-1(3))=cos(phi)="adjacent"/"hypotenuse"=1/sqrt10#
#sin(tan^-1(3))=sin(phi)="opposite"/"hypotenuse"=3/sqrt10#
Plugging these into the expression from earlier we get:
#=1/sqrt5(1/sqrt10)-2/sqrt5(3/sqrt10)#
Note that
#=(1-6)/(5sqrt2)=-1/sqrt2#
Explanation:
Let
Therefore
As
and hence
and
=
Hence,