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Question: The maximum value of \[\left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2}\...

The maximum value of (b+c−a2R)sin⁡A⋅sec⁡2A2\left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} is
A) 11
B) 32\dfrac{3}{2}
C) 28\dfrac{2}{8}
D) None of these

Explanation

Solution

Here, we will first use the law sines to convert sine function in terms of variables. Then we will use the trigonometric and algebraic identities to simplify the expression further and find the maximum value of the trigonometric equation. Trigonometric equation is an equation involving the trigonometric ratios.

Formula Used:
We will use the following formula:

  1. Law of Sines is given by the formula a=2Rsin⁡Aa = 2R\sin A; b=2Rsin⁡Bb = 2R\sin B;c=2Rsin⁡Cc = 2R\sin C;
  2. Trigonometric Identity: sec⁡2A=1cos⁡2A=11+cos⁡A2{\sec ^2}A = \dfrac{1}{{{{\cos }^2}A}} = \dfrac{1}{{\dfrac{{1 + \cos A}}{2}}}
  3. Law of cosines is given by the formula cos⁡A=b2+c2−a22bc\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}
  4. The difference of square of numbers is given by the algebraic identity (a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)

Complete step by step solution:
We are given a function (b+c−a2R)sin⁡A⋅sec⁡2A2\left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2}
Law of Sines is given by the formula a=2Rsin⁡Aa = 2R\sin A; b=2Rsin⁡Bb = 2R\sin B;c=2Rsin⁡Cc = 2R\sin C;
By using the law of sines, we have a=2Rsin⁡Aa = 2R\sin A
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(b+c−a2R)(a2R)⋅sec⁡2A2\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{b + c - a}}{{2R}}} \right)\left( {\dfrac{a}{{2R}}} \right) \cdot {\sec ^2}\dfrac{A}{2}
Trigonometric Identity: sec⁡2A=1cos⁡2A=11+cos⁡A2{\sec ^2}A = \dfrac{1}{{{{\cos }^2}A}} = \dfrac{1}{{\dfrac{{1 + \cos A}}{2}}}
By rewriting using the trigonometric identity, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(a(b+c−a)4R2)⋅1cos⁡2A2\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{a\left( {b + c - a} \right)}}{{4{R^2}}}} \right) \cdot \dfrac{1}{{{{\cos }^2}\dfrac{A}{2}}}
By using the trigonometric identity, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(a(b+c−a)4R2)⋅1(1+cos⁡A2)\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{a\left( {b + c - a} \right)}}{{4{R^2}}}} \right) \cdot \dfrac{1}{{\left( {\dfrac{{1 + \cos A}}{2}} \right)}}
By rewriting the equation, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(a(b+c−a)4R2)⋅21+cos⁡A\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{a\left( {b + c - a} \right)}}{{4{R^2}}}} \right) \cdot \dfrac{2}{{1 + \cos A}}
Law of cosines is given by the formula cos⁡A=b2+c2−a22bc\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}
By the law of cosines, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(a(b+c−a)4R2)⋅2(1+(b2+c2−a22bc))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{a\left( {b + c - a} \right)}}{{4{R^2}}}} \right) \cdot \dfrac{2}{{\left( {1 + \left( {\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)} \right)}}
By rewriting using the algebraic identity, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(2abc(b+c−a)2R2[(b+c)2−a2])\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{2abc\left( {b + c - a} \right)}}{{2{R^2}\left[ {{{\left( {b + c} \right)}^2} - {a^2}} \right]}}} \right)
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(2abc(b+c−a)2R2[(b+c)2−a2])\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{2abc\left( {b + c - a} \right)}}{{2{R^2}\left[ {{{\left( {b + c} \right)}^2} - {a^2}} \right]}}} \right)
The difference of square of numbers is given by the algebraic identity (a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)
By rewriting using the algebraic identity, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(2abc(b+c−a)2R2(b+c−a)(a+b+c))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{2abc\left( {b + c - a} \right)}}{{2{R^2}\left( {b + c - a} \right)\left( {a + b + c} \right)}}} \right)
By cancelling the like terms, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abcR2(a+b+c))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{{R^2}\left( {a + b + c} \right)}}} \right)
By using the law of sines, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abcR2(2Rsin⁡A+2Rsin⁡B+2Rsin⁡C))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{{R^2}\left( {2R\sin A + 2R\sin B + 2R\sin C} \right)}}} \right)
By taking the common factors, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abcR2⋅2R(sin⁡A+sin⁡B+sin⁡C))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{{R^2} \cdot 2R\left( {\sin A + \sin B + \sin C} \right)}}} \right)
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abcR2⋅2R(sin⁡A+sin⁡B+sin⁡C))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{{R^2} \cdot 2R\left( {\sin A + \sin B + \sin C} \right)}}} \right)
We know that A+B+C=πA + B + C = \pi , so we get sin⁡A+sin⁡B+sin⁡C=4sin⁡Asin⁡Bsin⁡C\sin A + \sin B + \sin C = 4\sin A\sin B\sin C
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abc2R3⋅4sin⁡Asin⁡Bsin⁡C)\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{2{R^3} \cdot 4\sin A\sin B\sin C}}} \right)
By rewriting using the law of sines, we get
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abc(2Rsin⁡A)(2Rsin⁡B)(2Rsin⁡C))\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{\left( {2R\sin A} \right)\left( {2R\sin B} \right)\left( {2R\sin C} \right)}}} \right)
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=(abcabc)\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = \left( {\dfrac{{abc}}{{abc}}} \right)
⇒(b+c−a2R)sin⁡A⋅sec⁡2A2=1\Rightarrow \left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} = 1
Therefore, the maximum value of (b+c−a2R)sin⁡A⋅sec⁡2A2\left( {\dfrac{{b + c - a}}{{2R}}} \right)\sin A \cdot {\sec ^2}\dfrac{A}{2} is 1.

Thus Option(A) is the correct answer.

Note:
We know that the law of sines is an equation which is used to find the length of the sides of the triangle. The law of cosines is useful for finding the third side of the triangle when we know only two sides of a triangle and the angle between the angles of a triangle. We should know that both the law of sines and law of cosines are used in finding the area of the triangle and length of the sides of the triangle.