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Question: If \[\sin \theta +\cos \theta =1\] then \[\sin \theta \cos \theta \] is equal to A. 0 B. \[\dfra...

If sinθ+cosθ=1\sin \theta +\cos \theta =1 then sinθcosθ\sin \theta \cos \theta is equal to
A. 0
B. 12\dfrac{1}{2}
C. 1
D. 12-\dfrac{1}{2}

Explanation

Solution

firstly squaring the given data on both the left hand side and right hand side and using the basic trigonometric identity that is sin2θ+cos2θ=1{{\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 sinθ+cosθ=1\sin \theta +\cos \theta =1
We have to find the value of sinθcosθ\sin \theta \cos \theta .
sinθ+cosθ=1\sin \theta +\cos \theta =1
Squaring on both left hand side and right hand side then we will get,
(sinθ+cosθ)2=(1)2{{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( 1 \right)}^{2}}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know the property (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab.
Applying the above mentioned property to simplify the expression we get,
sin2θ+cos2θ+2sinθcosθ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta =1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
We know the property sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
Using the above mentioned property to solve the expression we get,
1+2sinθcosθ=11+2\sin \theta \cos \theta =1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Subtracting with 1 on both sides we get,
2sinθcosθ=02\sin \theta \cos \theta =0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
Dividing with 2 on both sides we get,
sinθcosθ=0\sin \theta \cos \theta =0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
So the obtained value of sinθcosθ\sin \theta \cos \theta is equal to 0.

So, the correct answer is “Option A”.

Note: Use the property sin2θ+cos2θ=1{{\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.