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Question

If A,B,C are angle of a triangle ABC, then the value of the determinant ∣ ∣ ∣ ∣ ∣ ∣sinA2sinB2sinC2sin(A+B+C)sinB2cosA2cos(A+B+C)2tan(A+B+C)sinC2∣ ∣ ∣ ∣ ∣ ∣ is less than or equal to

A
12
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B
14
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C
18
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D
None of these
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Solution

The correct option is B 18
Since A,B,C are the angles of a triangle so by angle sum property of a triangle, A+B+C=π
Putting in this value in the given determinant, we get-

Δ=∣ ∣ ∣ ∣sinA2sinB2sinC2sin(π)sinB2cosA2cosπ2tan(π)sinC2∣ ∣ ∣ ∣

=∣ ∣ ∣ ∣sinA2sinB2sinC20sinB2cosA200sinC2∣ ∣ ∣ ∣
Expanding along R3 we get-

=sinC2(sinA2×sinB20)

=sinA2.sinB2.sinC2

Let sinA2.sinB2.sinC2k

Then, k is the maximum value of sinA2.sinB2.sinC2

Since, A+B+C=π

A+B+C2=π2A+B2+C2=π2

C2=π2A+B2

Thus, sinA2.sinB2.sinC2

=sinA2.sinB2.sin(π2A+B2)

=sinA2.sinB2.cos(A+B2)

Let f(A,B)=sinA2.sinB2.cos(A+B2)

Now, fA=12sinB2cos(A+B2)

and, fB=12sinA2cos(B+A2)

At the point of maxima of f, fA=fB=0

12sinB2cos(A+B2)=12sinA2cos(B+A2)=0

Considering A,B to be non zero, we have-

cos(A+B2)=cos(B+A2)=0

A+B2=B+A2=π2

Solving we get, A=B=π3 and thus, C=πAB=π3

Now, 2fA2=A(fA)

2fA2=A(12sinB2cos(A+B2))

2fA2=12sinB2sin(A+B2)

At A=B=π3,

2fA2=12sinπ6sin(π3+π6)=14

Again,

2fB2=B(fB)

2fB2=B(12sinA2cos(B+A2))

2fB2=12sinA2sin(B+A2)

At A=B=π3,

2fB2=12sinπ6sin(π3+π6)=14

And,

2fA.B=A(fB)

2fA.B=A(12sinA2cos(B+A2))

2fA.B=14cos(A+B)

Thus, at A=B=π3,

2fA.B=14cos(π3+π3)=14×13=112

Now, D=2fA2.2fB2(2fA.B)2=(14)(14)(112)2=1161144=118>0

And, 2fA2=14<0

Hence, A=B=C=π3 is a point of maxima.

Thus, the maximum value of sinA2.sinB2.sinC2=sinπ6.sinπ6.sinπ6=(12)2=18

Thus, sinA2.sinB2.sinC218

Hence, the correct answer is option C.

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