Question
Question: If \[\sin \theta +\cos \theta =1\] then \[\sin \theta \cos \theta \] is equal to A. 0 B. \[\dfra...
If sinθ+cosθ=1 then sinθcosθ is equal to
A. 0
B. 21
C. 1
D. −21
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. 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
We have to find the value of sinθcosθ.
sinθ+cosθ=1
Squaring on both left hand side and right hand side then we will get,
(sinθ+cosθ)2=(1)2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know the property (a+b)2=a2+b2+2ab.
Applying the above mentioned property to simplify the expression we get,
sin2θ+cos2θ+2sinθcosθ=1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
We know the property sin2θ+cos2θ=1.
Using the above mentioned property to solve the expression we get,
1+2sinθcosθ=1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Subtracting with 1 on both sides we get,
2sinθcosθ=0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)
Dividing with 2 on both sides we get,
sinθcosθ=0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
So the obtained value of sinθcosθis equal to 0.
So, the correct answer is “Option A”.
Note: Use the property sin2θ+cos2θ=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.