Question

If x and y are two sides of a square and $x = \left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) = \left( {1 - \sin A} \right)\left( {1 - \sin B} \right)\left( {1 - \sin C} \right)$ show that each side is equal to $ \pm \cos A\cos B\cos C$.

Answer 1

Hint:- In this question use the Pythagorean identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and also the property of square that all of its sides are equal.

Complete step-by-step solution -
Given that:
$\left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) = \left( {1 - \sin A} \right)\left( {1 - \sin B} \right)\left( {1 - \sin C} \right)$
Multiply both sides by $\left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right)$ we get:
$
  \left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) \times \left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) = \left( {1 - \sin A} \right)\left( {1 - \sin B} \right)\left( {1 - \sin C} \right) \times \left( {1 + \sin A} \right) \\
  \left( {1 + \sin B} \right)\left( {1 + \sin C} \right) \\
$
We know that $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
Equation can be now be written as: ${\left( {1 + \sin A} \right)^2}{\left( {1 + \sin B} \right)^2}{\left( {1 + \sin C} \right)^2} = \left( {1 - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}B} \right)\left( {1 - {{\sin }^2}C} \right)$
As we know that:
$ {{sin} ^{2}}\theta + {{cos} ^{2}}\theta = 1 \\
  \therefore {{cos} ^2}\theta = 1 - {{sin} ^{2}}\theta \\ $
Equation can be further written as: ${\left( {1 + \sin A} \right)^2}{\left( {1 + \sin B} \right)^2}{\left( {1 + \sin C} \right)^2} = {\cos ^2}A{\cos ^2}B{\cos ^2}C$
Therefore\[{\left[ {\left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right)} \right]^2} = {\left[ {\cos A\cos B\cos C} \right]^2}\]
Taking square roots both sides we get:
$\left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) = \pm \cos A\cos B\cos C$
Therefore $x = \left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right) = \left( {1 - \sin A} \right)\left( {1 - \sin B} \right)\left( {1 - \sin C} \right) = \pm \cos A\cos B\cos C$
We know that all sides of a square are equal
Hence proved that each side is equal to $ \pm \cos A\cos B\cos C$

Note:- In this question first we multiplied the given equation with $\left( {1 + \sin A} \right)\left( {1 + \sin B} \right)\left( {1 + \sin C} \right)$ and simplified it using algebraic formula $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ after that we applied the value of ${\cos ^2}\theta $ from the Pythagorean identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and simplified the equation and got the final equation which signifies that each side of square is equal to $ \pm \cos A\cos B\cos C$.