If in a triangle ABC `cosAcosB+sinAsinBsinC=1` then how will you prove that the triangle is right angled and isosceles?

Multiplying both sides by `2` in given equality `cosAcosB+sinAsinBsinC=1`, we get

`2cosAcosB+2sinAsinBsinC=2` or

`2cosAcosB+2sinAsinBsinC=(sin^2A+cos^2A)+(sin^2B+cos^2B)`

or `(cos^2A+cos^2B-2cosAcosB)+(sin^2A+sin^2B-2sinAsinB)+2sinAsinB-2sinAsinBsinC=0` or

or `(cosA-cosB)^2+(sinA-sinB)^2+2sinAsinB(1-sinC)=0`

Note that all three terms contained above are positive, as while first two terms are squares and hence positive, third term is positive as sine of angles `A` and `B` is positive (as they are less than `pi` being angles of a triangle) and `(1-sinC)` too will be positive as `sinC<1`.

But, their sum is zero and hence each term is equal to zero, i.e.

`cosA-cosB=0`, `sinA-sinB=0` and `1-sinC=0` or

`cosA=cosB`, `sinA=sinB` and `1-sinC=0`

i.e. `A=B` and `sinC=1` i.e. `C=pi/2`

Hence the triangle is isosceles and right angled.

Given
`cosAcosB+sinAsinBsinC=1`
`=>cosAcosB+sinAsinB-sinAsinB+sinAsinBsinC=1`

`=>cos(A-B)-sinAsinB(1-sinC)=1`

`=>1-cos(A-B)+sinAsinB(1-sinC)=0`
`=>2sin^2((A-B)/2)+sinAsinB(1-sinC)=0`

Now in above relation the first term being squared quantity will be positive.In the second term A,B and C all are less than
`180^@` but greater than zero.
So sinA ,sinB and sinC all are positive and less than 1.So the 2nd term as a whole is positive.
But RHS=0.
It is only possible iff each term becomes zero.

When `2sin^2((A-B)/2)=0`
then`A=B`

and when 2nd term=0 then
`sinAsinB(1-sinC)=0`

0< A and B <180
`=>sinA !=0and sinB!=0`

So `1-sinC=0=>C=pi/2`

So in triangle ABC
`A=B and C=pi/2->"the triangle is right angled and isosceles"`