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

Hint: firstly squaring the given data on both the left hand side and right hand side and using the basic trigonometric identity that is \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By substituting this value and doing basic mathematical operations like addition etc then we will get the required value.

Complete step-by-step answer:
Given that \[\sin \theta +\cos \theta =1\]
We have to find the value of \[\sin \theta \cos \theta \].
 \[\sin \theta +\cos \theta =1\]
Squaring on both left hand side and right hand side then we will get,
 \[{{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( 1 \right)}^{2}}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know the property \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
Applying the above mentioned property to simplify the expression we get,
 \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =1\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
We know the property \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
Using the above mentioned property to solve the expression we get,
 \[1+2\sin \theta \cos \theta =1\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Subtracting with 1 on both sides we get,
 \[2\sin \theta \cos \theta =0\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
Dividing with 2 on both sides we get,
 \[\sin \theta \cos \theta =0\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
So the obtained value of \[\sin \theta \cos \theta \]is equal to 0.

So, the correct answer is “Option A”.

Note: Use the property \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify the given problem. The possible error that you may encounter can be the wrong substitution of the trigonometric property. Calculations should be carried out carefully.