Question

If A and B are complementary angles then ${\sin ^2}A + {\sin ^2}B =$A. $0$B. $\dfrac{1}{2}$C. $2$D. None of these

Hint: First of all, write the two angles in terms of one angle as their sum is equal to ${90^0}$ (since complementary angles). Then find the result by using trigonometric identities. So, use this concept to reach the solution of the problem.

Complete step-by-step solution -
Given that angles A and B are complementary angles.
So, $A + B = {90^0}$
$\Rightarrow A = {90^0} - B.......................................................\left( 1 \right)$
To find ${\sin ^2}A + {\sin ^2}B =$
$\Rightarrow {\sin ^2}\left( {{{90}^0} - B} \right) + {\sin ^2}B$
We know that $\sin \left( {{{90}^0} - x} \right) = \cos x$
So, we have
$\Rightarrow {\cos ^2}B + {\sin ^2}B$
We know that ${\cos ^2}x + {\sin ^2}x = 1$
By using the above formula, we get
$\Rightarrow {\cos ^2}B + {\sin ^2}B = 1$
Therefore, ${\sin ^2}A + {\sin ^2}B = 1$
Thus, the correct option is D. None of these

Note: We can write $\sin \left( {{{90}^0} - x} \right)$ as $\cos x$ because their values do not change and they lie in the same quadrant. Two angles are said to be complementary angles when the sum of their individual angles is equal to ${90^0}$.