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

Jun 27, 2016

#### Explanation:

Multiplying both sides by $2$ in given equality $\cos A \cos B + \sin A \sin B \sin C = 1$, we get

$2 \cos A \cos B + 2 \sin A \sin B \sin C = 2$ or

$2 \cos A \cos B + 2 \sin A \sin B \sin C = \left({\sin}^{2} A + {\cos}^{2} A\right) + \left({\sin}^{2} B + {\cos}^{2} B\right)$

or $\left({\cos}^{2} A + {\cos}^{2} B - 2 \cos A \cos B\right) + \left({\sin}^{2} A + {\sin}^{2} B - 2 \sin A \sin B\right) + 2 \sin A \sin B - 2 \sin A \sin B \sin C = 0$ or

or ${\left(\cos A - \cos B\right)}^{2} + {\left(\sin A - \sin B\right)}^{2} + 2 \sin A \sin B \left(1 - \sin C\right) = 0$

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 $\left(1 - \sin C\right)$ too will be positive as $\sin C < 1$.

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

$\cos A - \cos B = 0$, $\sin A - \sin B = 0$ and $1 - \sin C = 0$ or

$\cos A = \cos B$, $\sin A = \sin B$ and $1 - \sin C = 0$

i.e. $A = B$ and $\sin C = 1$ i.e. $C = \frac{\pi}{2}$

Hence the triangle is isosceles and right angled.

Jun 27, 2016

as follows

#### Explanation:

Given
$\cos A \cos B + \sin A \sin B \sin C = 1$
$\implies \cos A \cos B + \sin A \sin B - \sin A \sin B + \sin A \sin B \sin C = 1$

$\implies \cos \left(A - B\right) - \sin A \sin B \left(1 - \sin C\right) = 1$

$\implies 1 - \cos \left(A - B\right) + \sin A \sin B \left(1 - \sin C\right) = 0$
$\implies 2 {\sin}^{2} \left(\frac{A - B}{2}\right) + \sin A \sin B \left(1 - \sin C\right) = 0$

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
${180}^{\circ}$ 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 $2 {\sin}^{2} \left(\frac{A - B}{2}\right) = 0$
then$A = B$

and when 2nd term=0 then
$\sin A \sin B \left(1 - \sin C\right) = 0$

0< A and B <180
$\implies \sin A \ne 0 \mathmr{and} \sin B \ne 0$

So $1 - \sin C = 0 \implies C = \frac{\pi}{2}$

So in triangle ABC
$A = B \mathmr{and} C = \frac{\pi}{2} \to \text{the triangle is right angled and isosceles}$