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Question: Given that \[\tan A\], \[\tan B\] are the roots of the equation \[{x^2} - bx + c\], the value of \[{...

Given that tanA\tan A, tanB\tan B are the roots of the equation x2bx+c{x^2} - bx + c, the value of sin2(A+B){\sin ^2}\left( {A + B} \right) is
1. b(b+c)2\dfrac{b}{{{{\left( {b + c} \right)}^2}}}
2. b2b2+c2\dfrac{{{b^2}}}{{{b^2} + {c^2}}}
3. b2c2+(1b2)\dfrac{{{b^2}}}{{{c^2} + \left( {1 - {b^2}} \right)}}
4. b2b2+(1c)2\dfrac{{{b^2}}}{{{b^2} + {{\left( {1 - c} \right)}^2}}}

Explanation

Solution

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(A+B)=tanA+tanB1tanAtanB\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}}
2. 1+cot2θ=cosec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta
3. sinθ=1cosecθ\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. sin2(A+B){\sin ^2}\left( {A + B} \right)
It is given that tanA\tan A and tanB\tan B are the roots of the given equation x2bx+c{x^2} - bx + c.
Now, we will find the sum and the product of the roots.
We know that if x2+bx+c{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 xx to the coefficient of the term x2{x^2} and the product of the roots of the equation is equal to the ratio of the constant term ccto the coefficient of the term x2{x^2}.
Using this concept, we can write
tanA+tanB=b1=b\tan A + \tan B = - \dfrac{{ - b}}{1} = b …………. (1)\left( 1 \right)
tanAtanB=c1=c\tan A \cdot \tan B = \dfrac{c}{1} = c ……………….. (2)\left( 2 \right)
We know from the trigonometry identities that tan(A+B)=tanA+tanB1tanAtanB\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}}.
Now, we will substitute the values from equation (1)\left( 1 \right) and equation (2)\left( 2 \right) in the above formula. Therefore, we get
tan(A+B)=b1c\tan \left( {A + B} \right) = \dfrac{b}{{1 - c}}
Squaring terms on both sides, we get
tan2(A+B)=b2(1c)2\Rightarrow {\tan ^2}\left( {A + B} \right) = \dfrac{{{b^2}}}{{{{\left( {1 - c} \right)}^2}}}
Using the formula tanθ=1cotθ\tan \theta = \dfrac{1}{{\cot \theta }}, we can write above equation as:
1cot2(A+B)=b2(1c)2\Rightarrow \dfrac{1}{{{{\cot }^2}\left( {A + B} \right)}} = \dfrac{{{b^2}}}{{{{\left( {1 - c} \right)}^2}}}
On cross multiplication, we get
cot2(A+B)=(1c)2b2\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
1+cot2(A+B)=1+(1c)2b2\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
1+cot2(A+B)=b2+(1c)2b2\Rightarrow 1 + {\cot ^2}\left( {A + B} \right) = \dfrac{{{b^2} + {{\left( {1 - c} \right)}^2}}}{{{b^2}}}
Now using the formula 1+cot2θ=cosec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta , we get

cosec2(A+B)=b2+(1c)2b2 \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 sinθ=1cosecθ\sin \theta = \dfrac{1}{{\cos ec\theta }}.
Using this formula, we get
1sin2(A+B)=b2+(1c)2b2\Rightarrow \dfrac{1}{{{{\sin }^2}\left( {A + B} \right)}} = \dfrac{{{b^2} + {{\left( {1 - c} \right)}^2}}}{{{b^2}}}
On cross multiplication, we get
sin2(A+B)=b2b2+(1c)2\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.