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Question: Assertion(A): \({\cos ^2}\theta + {\cos ^2}({60^ \circ } - \theta ) + {\cos ^2}({60^ \circ } - \th...

Assertion(A):
cos⁡2θ+cos⁡2(60∘−θ)+cos⁡2(60∘−θ)=32{\cos ^2}\theta + {\cos ^2}({60^ \circ } - \theta ) + {\cos ^2}({60^ \circ } - \theta ) = \dfrac{3}{2}
Reason(R):
sin⁡α−sin⁡(120∘−α)+sin⁡(120∘+α)=0\sin \alpha - \sin ({120^ \circ } - \alpha ) + \sin ({120^ \circ } + \alpha ) = 0
Note: xxin cos⁡x\cos xis represented in degrees.
A.{\text{A}}{\text{.}}The assertion is true, Reason is true and Reason is correct explanation of Assertion.
B.{\text{B}}{\text{.}}The assertion is true. Reason is true and Reason is not the correct explanation of Assertion.
C.{\text{C}}{\text{.}}Assertion is true. Reason is false.
D.{\text{D}}{\text{.}}Assertion is false. Reason is true.

Explanation

Solution

Hint:In this question we first evaluate the Assertion and then Reason. Using Trigonometric Properties we have to check if LHS is equal to RHS or not. If both the assertion and reason are true then we have to check if reason explains the assertion or not i.e. the solution of reason is used in solving the assertion or not.

Complete step-by-step answer:
Initially, considering the Assertion(A):
cos⁡2θ+cos⁡2(60∘−θ)+cos⁡2(60∘−θ)=32{\cos ^2}\theta + {\cos ^2}({60^ \circ } - \theta ) + {\cos ^2}({60^ \circ } - \theta ) = \dfrac{3}{2} eq.1
Considering the LHS of eq.1
⇒cos⁡2θ+cos⁡2(60∘−θ)+cos⁡2(60∘ + θ)\Rightarrow {\cos ^2}\theta + {\cos ^2}({60^ \circ } - \theta ) + {\cos ^2}({60^ \circ }{\text{ + }}\theta )
Now add and subtract 2cos⁡(60∘−θ)cos⁡(60∘+θ)2\cos ({60^ \circ } - \theta )\cos ({60^ \circ } + \theta ) to the above equation, we get
⇒cos⁡2θ+cos⁡2(60∘−θ)+cos⁡2(60∘ + θ)+2cos⁡(60∘−θ)cos⁡(60∘+θ)−2cos⁡(60∘−θ)cos⁡(60∘+θ) ⇒cos⁡2θ+(cos⁡(60∘−θ)+cos⁡(60∘ + θ))2−2cos⁡(60∘−θ)cos⁡(60∘+θ) eq.2  \Rightarrow {\cos ^2}\theta + {\cos ^2}({60^ \circ } - \theta ) + {\cos ^2}({60^ \circ }{\text{ + }}\theta ) + 2\cos ({60^ \circ } - \theta )\cos ({60^ \circ } + \theta ) - 2\cos ({60^ \circ } - \theta )\cos ({60^ \circ } + \theta ) \\\ \Rightarrow {\cos ^2}\theta + {(\cos ({60^ \circ } - \theta ) + \cos ({60^ \circ }{\text{ + }}\theta ))^2} - 2\cos ({60^ \circ } - \theta )\cos ({60^ \circ } + \theta ){\text{ eq}}{\text{.2}} \\\
Now using the property of cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos (a + b) = \cos a\cos b - \sin a\sin b . We can write eq.2 as
⇒cos⁡2θ+(2cos⁡60∘cosθ)2−2cos⁡(60∘−θ)cos⁡(60∘+θ) \Rightarrow {\cos ^2}\theta + {(2\cos {60^ \circ }{\text{cos}}\theta )^2} - 2\cos ({60^ \circ } - \theta )\cos ({60^ \circ } + \theta ){\text{ }} eq.3
Now using the property of
cos⁡(a+b)+cos⁡(a−b)=2cos⁡(a+b2)sin⁡(a+b2)\cos (a + b) + \cos (a - b) = 2\cos (\dfrac{{a + b}}{2})\sin (\dfrac{{a + b}}{2}) we can rewrite eq.3 as
⇒cos⁡2θ+(2cos⁡60∘cosθ)2−(cos⁡(120∘) + cos2θ) ⇒cos⁡2θ+ cos⁡2θ −(cos2θ−12)  cos60∘=12 ⇒2cos⁡2θ+−cos⁡2θ+12  \Rightarrow {\cos ^2}\theta + {(2\cos {60^ \circ }{\text{cos}}\theta )^2} - (\cos ({120^ \circ }{\text{) + cos2}}\theta {\text{)}} \\\ \Rightarrow {\cos ^2}\theta + {\text{ }}{\cos ^2}\theta {\text{ }} - ({\text{cos2}}\theta - \dfrac{1}{2}){\text{ \\{ cos6}}{0^ \circ } = \dfrac{1}{2}\\} \\\ \Rightarrow 2{\cos ^2}\theta + - \cos 2\theta + \dfrac{1}{2} \\\
We know,cos2θ=2cos⁡2θ−1{\text{cos2}}\theta = 2{\cos ^2}\theta - 1 then we can write above equation as
⇒2cos⁡2θ−(2cos⁡2θ−1)+12 ⇒32  \Rightarrow 2{\cos ^2}\theta - (2{\cos ^2}\theta - 1) + \dfrac{1}{2} \\\ \Rightarrow \dfrac{3}{2} \\\
We can clearly see that LHS=RHS of Assertion.
Hence it is True.
Now consider the Reason(R):
sin⁡α−sin⁡(120∘−α)+sin⁡(120∘+α)=0 eq.4\sin \alpha - \sin ({120^ \circ } - \alpha ) + \sin ({120^ \circ } + \alpha ) = 0{\text{ eq}}{\text{.4}}
Now, considering the LHS of eq.4, we get
We know , sin⁡(a+b)−sin⁡(a−b)=2cos⁡asin⁡b\sin (a + b) - \sin (a - b) = 2\cos a\sin b
Using above property we can rewrite eq.4 as
⇒sin⁡α+2cos⁡(120∘)sin⁡α ⇒sin⁡α−sin⁡α  cos⁡(120∘)=−12 ⇒0  \Rightarrow \sin \alpha + 2\cos ({120^ \circ })\sin \alpha \\\ \Rightarrow \sin \alpha - \sin \alpha {\text{ \\{ }}\cos ({120^ \circ }) = - \dfrac{1}{2}\\} \\\ \Rightarrow 0 \\\
We can clearly see that LHS=RHS of Reason.
Hence it is True.
As we see, Assertion(A) can be explained without the need of Reason(R).
So, option B is true. The assertion is true. Reason is true and Reason is not the correct explanation of Assertion.

Note: Whenever you get this type of question the key concept to solve this is to evaluate the LHS of both the Assertion and Reason and check if they are equal or not. If they both are true then check one thing that is Reason’s concept is used in Assertion or not.