Question

If $\sin \theta + \cos \theta = 1$, then $(\sin \theta \cos \theta )$ is equal to
(a) $0$
(b )$\dfrac{1}{2}$
(c) $1$
(d) $- \dfrac{1}{2}$

Answer 1

As we know that the above given question is related to trigonometric expression, sine and cosine are trigonometric ratios. Here we have to find the value using trigonometric identity or formulae. Here we will square both the left hand side and the right hand side to get the required value. We can also use the trigonometric identity of sine and cosine sum formula

Complete step by step solution:
As per the given question we have $\sin \theta + \cos \theta = 1$, and we have to find the value of $\sin \theta \cos \theta$.
We will first square both the sides of the given equation and we get: ${(\sin \theta + \cos \theta)^2} = {1^2}$. We know that ${(a + b)^2} = {a^2} + 2ab + {b^2}$. Expanding the equation we get,
${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 1$.
Also we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$, so by substituting the value we get: $1 + 2\sin \theta \cos \theta = 1 \Rightarrow 2\sin \theta \cos \theta = 1 - 1$.
It gives us $\sin \theta \cos \theta = \dfrac{0}{2} = 0$.
Hence the correct option is (a) $0$.

Note: Before solving such a question we should be fully aware of the trigonometric identities, ratios and their formulas. We should remember them as we need to use them in solving questions like this. We should be careful while doing the calculation because if there is mistake in calculation, we might get the wrong answer.