Answer

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**Hint:**Here we need to find the value of the given trigonometric expression. For that, we will first find the sum of the roots of the given quadratic equation. Then we will find the product of the given roots. We will then use the basic trigonometric formulas to proceed further and then we will substitute the value of sum and the product of the roots obtained in the formula. After simplifying the terms using the mathematical operations, we will get the required value of the given trigonometric expression.

**Formula used:**

We will use the following formulas:

1. \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}}\]

2. \[1 + {\cot ^2}\theta = \cos e{c^2}\theta \]

3. \[\sin \theta = \dfrac{1}{{\cos ec\theta }}\]

**Complete step by step solution:**

Here we need to find the value of the given trigonometric expression i.e. \[{\sin ^2}\left( {A + B} \right)\]

It is given that \[\tan A\] and \[\tan B\] are the roots of the given equation \[{x^2} - bx + c\].

Now, we will find the sum and the product of the roots.

We know that if \[{x^2} + bx + c\] is the quadratic equation then the sum of the roots of the equation is equal to the negative of the ratio of the coefficient of the term \[x\] to the coefficient of the term \[{x^2}\] and the product of the roots of the equation is equal to the ratio of the constant term \[c\]to the coefficient of the term \[{x^2}\].

Using this concept, we can write

\[\tan A + \tan B = - \dfrac{{ - b}}{1} = b\] …………. \[\left( 1 \right)\]

\[\tan A \cdot \tan B = \dfrac{c}{1} = c\] ……………….. \[\left( 2 \right)\]

We know from the trigonometry identities that \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}}\].

Now, we will substitute the values from equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] in the above formula. Therefore, we get

\[\tan \left( {A + B} \right) = \dfrac{b}{{1 - c}}\]

Squaring terms on both sides, we get

\[ \Rightarrow {\tan ^2}\left( {A + B} \right) = \dfrac{{{b^2}}}{{{{\left( {1 - c} \right)}^2}}}\]

Using the formula \[\tan \theta = \dfrac{1}{{\cot \theta }}\], we can write above equation as:

\[ \Rightarrow \dfrac{1}{{{{\cot }^2}\left( {A + B} \right)}} = \dfrac{{{b^2}}}{{{{\left( {1 - c} \right)}^2}}}\]

On cross multiplication, we get

\[ \Rightarrow {\cot ^2}\left( {A + B} \right) = \dfrac{{{{\left( {1 - c} \right)}^2}}}{{{b^2}}}\]

Now, we will add 1 to both sides. So, we get

\[ \Rightarrow 1 + {\cot ^2}\left( {A + B} \right) = 1 + \dfrac{{{{\left( {1 - c} \right)}^2}}}{{{b^2}}}\]

Taking LCM on the right hand side, we get

\[ \Rightarrow 1 + {\cot ^2}\left( {A + B} \right) = \dfrac{{{b^2} + {{\left( {1 - c} \right)}^2}}}{{{b^2}}}\]

Now using the formula \[1 + {\cot ^2}\theta = \cos e{c^2}\theta \], we get

\[ \Rightarrow \cos e{c^2}\left( {A + B} \right) = \dfrac{{{b^2} + {{\left( {1 - c} \right)}^2}}}{{{b^2}}}\]

We know from the trigonometry identities that \[\sin \theta = \dfrac{1}{{\cos ec\theta }}\].

Using this formula, we get

\[ \Rightarrow \dfrac{1}{{{{\sin }^2}\left( {A + B} \right)}} = \dfrac{{{b^2} + {{\left( {1 - c} \right)}^2}}}{{{b^2}}}\]

On cross multiplication, we get

\[ \Rightarrow {\sin ^2}\left( {A + B} \right) = \dfrac{{{b^2}}}{{{b^2} + {{\left( {1 - c} \right)}^2}}}\]

**Hence, the correct option is option 4.**

**Note:**

Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’ and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.

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