Question

If in $\Delta ABC$, ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1$ then the $\Delta ABC$ is a right-angled.

Answer 1

Hint: To solve the question, we have to apply the appropriate trigonometric identities to simplify the given expression. To solve the question further, apply the concept of the values of angles of trigonometric functions to the simplified expression. Thus, we can obtain the connection between the solution of the given expression and the properties of a right-angled triangle.

Complete step by step answer:
Let A, B, C be the angles of the triangle ABC.

The given expression is ${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1$

${{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C-1=0$

By substituting the trigonometric identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the above equation, we get

${{\cos }^{2}}A+{{\cos }^{2}}B-{{\sin }^{2}}C=0$

We know that the formula $\cos (\alpha +\beta )\cos (\alpha -\beta )={{\cos }^{2}}\alpha -{{\sin }^{2}}\beta$. By substituting this formula in the above equation, we get

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

We know the sum of all angles of triangle is equal to ${{180}^{0}}$

$A+B+C={{180}^{0}}$

$\Rightarrow B+C={{180}^{0}}-A$ ……(1)

By substituting this formula in the above equation, we get

${{\cos }^{2}}A+\cos ({{180}^{0}}-A)\cos (B-C)=0$

We know that $\cos \left( {{180}^{0}}-x \right)=-\cos x$. Thus, we get

${{\cos }^{2}}A-\cos (A)\cos (B-C)=0$

$\cos A\left( -\cos (A)+\cos (B-C) \right)=0$

By substituting the equation (1) in the above equation, we get

$\cos A\left( -\cos \left( {{180}^{0}}-(B+C) \right)+\cos (B-C) \right)=0$
By applying the above mentioned property, we get

$\cos A\left( -\left( -\cos \left( B+C \right) \right)+\cos \left( B-C \right) \right)=0$

$\cos A\left( \cos \left( B+C \right)+\cos \left( B-C \right) \right)=0$

We know that the formula $\cos (\alpha +\beta )+\cos (\alpha -\beta )=2\cos \alpha \cos \beta$. By substituting this formula in the above equation, we get

$\cos A\left( 2\cos B\cos C \right)=0$

$2\cos A\cos B\cos C=0$

$\cos A\cos B\cos C=0$

The above equation is true when either of the terms of the equation is equal to 0. The value of $\cos {{90}^{0}}$ is equal to 0.

Thus, either of the angles A, B, C is equal to ${{90}^{0}}$ which implies that it is right-angled.

Hence, proved

Note: The possibility of mistake can be not applying the appropriate formula of trigonometric identities to ease the procedure of solving. The other possibility of mistake is not applying the concept that $\cos {{90}^{0}}$ is equal 0, to the expression solved till it is equal to the product of cosine angles of triangle ABC.