
If \[m\tan \left( {\theta - 30} \right) = n\tan \left( {\theta + 120} \right)\] , then \[\dfrac{{m + n}}{{m - n}}\] is
A. \[2\cos 2\theta \]
B. \[\cos 2\theta \]
C. \[2\sin 2\theta \]
D. \[\sin 2\theta \]
Answer
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Hint: In this question, we have to find the value of \[\dfrac{{m + n}}{{m - n}}\]. Firstly, we will rearrange the terms by taking variables on one side and trigonometric functions on another side. Then, apply the componendo and dividendo rule and simplify the expression and further use the trigonometric value of a particular angle to find the value of \[\dfrac{{m + n}}{{m - n}}\].
Formula used: The formula used in this question is shown below;
1. Componendo and dividendo rule that is if \[\dfrac{{a}}{{b}} = \dfrac{{c}}{{d}}\] then \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\]
2. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
3. \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\]
4. \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\]
5. \[\sin \left( {180 - x} \right) = \sin x\]
6. \[\sin \left( {\theta + 90} \right) = \cos \theta \]
Complete step-by-step solution:
We are given that
\[m\tan \left( {\theta - 30} \right) = n\tan \left( {\theta + 120} \right)\]
Firstly, we will rearrange the given equation by taking variables on left-hand side and trigonometric function on right-hand side, we get
\[\dfrac{m}{n} = \dfrac{{\tan \left( {\theta + 120} \right)}}{{\tan \left( {\theta - 30} \right)}}\]
Let us assume \[\theta + 120 = A\] and \[\theta - 30 = B\].
Now, substitute this in the above equation, we get
\[\dfrac{m}{n} = \dfrac{{\tan A}}{{\tan B}}\] ……(1)
Further, we will apply the componendo and dividendo rule that is if \[\dfrac{{a}}{{b}} = \dfrac{{c}}{{d}}\] then \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\].
Here, \[a\] is \[m,b\] is \[n,c\] is \[\tan A\] and \[d\] is \[\tan B\] and substitute the values in equation (1), we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\tan A + \tan B}}{{\tan A - \tan B}}\]
Furthermore, we will apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]in the above equation where \[\theta \] is \[A\] and \[B\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\dfrac{{\sin A}}{{\cos A}} + \dfrac{{\sin B}}{{\cos B}}}}{{\dfrac{{\sin A}}{{\cos A}} - \dfrac{{\sin B}}{{\cos B}}}}\]
Now, we will simplify the above expression by taking LCM, we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}}}{{\dfrac{{\sin A\cos B - \sin A\sin B}}{{\cos A\cos B}}}}\\\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\sin A\cos B - \sin A\sin B}}\end{array}\]
Further, we will use the formula \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\] and \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\] on the above expression where \[x\] is \[A\] and \[y\] is \[B\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {A + B} \right)}}{{\sin \left( {A - B} \right)}}\]
Furthermore, we will resubstitute \[\theta + 120 = A\] and \[\theta - 30 = B\], we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {\theta + 120 + \theta - 30} \right)}}{{\sin \left( {\theta + 120 - \theta + 30} \right)}}\\\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {150} \right)}}\end{array}\]
Now, we will rewrite \[\sin \left( {150} \right)\] as \[\sin \left( {180 - 30} \right)\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {180 - 30} \right)}}\]
Further, we will apply the formula \[\sin \left( {180 - x} \right) = \sin x\] where \[x\] is \[30\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {30} \right)}}\]
Furthermore, we will apply the formula \[\sin \left( {\theta + 90} \right) = \cos \theta \], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\cos 2\theta }}{{\sin \left( {30} \right)}}\]
As we know \[\sin \left( {30} \right) = \dfrac{1}{2}\], we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\cos 2\theta }}{{\dfrac{1}{2}}}\\\dfrac{{m + n}}{{m - n}} = 2\cos 2\theta \end{array}\]
Hence, option A is correct
Note: In this type of question, we should remember the componendo and dividendo rule and know how to use them. We should also remember the addition and subtraction properties of the trigonometry function and also remember what happened when the angle is changed by the right angle.
Formula used: The formula used in this question is shown below;
1. Componendo and dividendo rule that is if \[\dfrac{{a}}{{b}} = \dfrac{{c}}{{d}}\] then \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\]
2. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
3. \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\]
4. \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\]
5. \[\sin \left( {180 - x} \right) = \sin x\]
6. \[\sin \left( {\theta + 90} \right) = \cos \theta \]
Complete step-by-step solution:
We are given that
\[m\tan \left( {\theta - 30} \right) = n\tan \left( {\theta + 120} \right)\]
Firstly, we will rearrange the given equation by taking variables on left-hand side and trigonometric function on right-hand side, we get
\[\dfrac{m}{n} = \dfrac{{\tan \left( {\theta + 120} \right)}}{{\tan \left( {\theta - 30} \right)}}\]
Let us assume \[\theta + 120 = A\] and \[\theta - 30 = B\].
Now, substitute this in the above equation, we get
\[\dfrac{m}{n} = \dfrac{{\tan A}}{{\tan B}}\] ……(1)
Further, we will apply the componendo and dividendo rule that is if \[\dfrac{{a}}{{b}} = \dfrac{{c}}{{d}}\] then \[\dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}\].
Here, \[a\] is \[m,b\] is \[n,c\] is \[\tan A\] and \[d\] is \[\tan B\] and substitute the values in equation (1), we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\tan A + \tan B}}{{\tan A - \tan B}}\]
Furthermore, we will apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]in the above equation where \[\theta \] is \[A\] and \[B\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\dfrac{{\sin A}}{{\cos A}} + \dfrac{{\sin B}}{{\cos B}}}}{{\dfrac{{\sin A}}{{\cos A}} - \dfrac{{\sin B}}{{\cos B}}}}\]
Now, we will simplify the above expression by taking LCM, we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\dfrac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}}}{{\dfrac{{\sin A\cos B - \sin A\sin B}}{{\cos A\cos B}}}}\\\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin A\cos B + \cos A\sin B}}{{\sin A\cos B - \sin A\sin B}}\end{array}\]
Further, we will use the formula \[\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\] and \[\sin \left( {x - y} \right) = \sin x\cos y - \cos x\sin y\] on the above expression where \[x\] is \[A\] and \[y\] is \[B\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {A + B} \right)}}{{\sin \left( {A - B} \right)}}\]
Furthermore, we will resubstitute \[\theta + 120 = A\] and \[\theta - 30 = B\], we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {\theta + 120 + \theta - 30} \right)}}{{\sin \left( {\theta + 120 - \theta + 30} \right)}}\\\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {150} \right)}}\end{array}\]
Now, we will rewrite \[\sin \left( {150} \right)\] as \[\sin \left( {180 - 30} \right)\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {180 - 30} \right)}}\]
Further, we will apply the formula \[\sin \left( {180 - x} \right) = \sin x\] where \[x\] is \[30\], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\sin \left( {2\theta + 90} \right)}}{{\sin \left( {30} \right)}}\]
Furthermore, we will apply the formula \[\sin \left( {\theta + 90} \right) = \cos \theta \], we get
\[\dfrac{{m + n}}{{m - n}} = \dfrac{{\cos 2\theta }}{{\sin \left( {30} \right)}}\]
As we know \[\sin \left( {30} \right) = \dfrac{1}{2}\], we get
\[\begin{array}{l}\dfrac{{m + n}}{{m - n}} = \dfrac{{\cos 2\theta }}{{\dfrac{1}{2}}}\\\dfrac{{m + n}}{{m - n}} = 2\cos 2\theta \end{array}\]
Hence, option A is correct
Note: In this type of question, we should remember the componendo and dividendo rule and know how to use them. We should also remember the addition and subtraction properties of the trigonometry function and also remember what happened when the angle is changed by the right angle.
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