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Question: The equations of \[{L_1}\] and \[{L_2}\] are \[y = mx\] and \[y = nx\], respectively. Suppose \[{L_1...

The equations of L1{L_1} and L2{L_2} are y=mxy = mx and y=nxy = nx, respectively. Suppose L1{L_1} makes twice as large of an angle with the horizontal (measured counterclockwise from the positive –axis) as does L2{L_2} and that L1{L_1} has 4 times the slope of L2{L_2}. If L1{L_1} is not horizontal, then the value of the product(mn){\text{(mn)}} equals.
A. 22\dfrac{{\sqrt 2 }}{2}
B. 22 - \dfrac{{\sqrt 2 }}{2}
C. 22
D. 2 - 2

Explanation

Solution

First, we will take m=tan2θm = \tan 2\theta and n=tanθn = \tan \theta and then use the property tan2θ=2tanθ1tan2θ\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} to simplify the values of mm and nn. Then substitute the obtained values in the product mnmn.

Complete step by step solution:
Given that L1{L_1} is y=mxy = mx and L2{L_2} is y=nxy = nx.

We know that the slope of L1{L_1} is mm and L2{L_2} is from the above equations.

Since it is given that the L1{L_1} has 4 times the slope of L2{L_2}, m=4nm = 4n.

Take m=tan2θm = \tan 2\theta and n=tanθn = \tan \theta , we get

m=4nm = 4n

We will use the property of tangential function, that is, tan2θ=2tanθ1tan2θ\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}, where θ\theta is the angle.

Using the above property in the above equation m=tan2θm = \tan 2\theta , we get

m=2tanθ1tan2θm = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}

Taking n=tanθn = \tan \theta in the above equation, we get

m=2n1n2m = \dfrac{{2n}}{{1 - {n^2}}}

Substituting this value of mm in the given equation m=4nm = 4n, we get
4n=2n1n24n = \dfrac{{2n}}{{1 - {n^2}}}

Dividing the above equation by 2n2n on each of the sides, we get

2=11n2 \Rightarrow 2 = \dfrac{1}{{1 - {n^2}}}

Cross-multiplying the above equation, we get

2(1n2)=1 22n2=1  \Rightarrow 2\left( {1 - {n^2}} \right) = 1 \\\ \Rightarrow 2 - 2{n^2} = 1 \\\

Subtracting the above equation by 2 on both sides, we get

22n22=12 2n22=12 n2=12  \Rightarrow 2 - 2{n^2} - 2 = 1 - 2 \\\ \Rightarrow \dfrac{{ - 2{n^2}}}{{ - 2}} = \dfrac{{ - 1}}{{ - 2}} \\\ \Rightarrow {n^2} = \dfrac{1}{2} \\\

Dividing the above equation by 2 - 2 on each of the sides, we get

Taking square root in the above equation on both sides, we get

n=±12 \Rightarrow n = \pm \dfrac{1}{{\sqrt 2 }}

Since we know that the slope of a line can never be negative, so the value 12 - \dfrac{1}{{\sqrt 2 }} is discarded.

Substituting the positive value of nn in the given equation m=4nm = 4n, we get

m=42m = \dfrac{4}{{\sqrt 2 }}

Now we will find the product mnmn from these values of mm and nn.

mn=42×12 mn=42 mn=2  mn = \dfrac{4}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow mn = \dfrac{4}{2} \\\ \Rightarrow mn = 2 \\\

Thus, the product mnmn equals 2.

Hence, the correct option is C.

Note:
In this question, we can also solve it by taking the angle made by y=mxy = mx with positive direction of –axis is tan1m{\tan ^{ - 1}}m and the angle made by line by y=nxy = nx is tan1n{\tan ^{ - 1}}n.
Now, we will take tan1m=2tan1n{\tan ^{ - 1}}m = 2{\tan ^{ - 1}}n.

We will use the property of tangential function, that is, 2tan1a=tan12a1a22{\tan ^{ - 1}}a = {\tan ^{ - 1}}\dfrac{{2a}}{{1 - {a^2}}}.

Using the above property in our assumed equation tan1m=2tan1n{\tan ^{ - 1}}m = 2{\tan ^{ - 1}}n, we get

tan1m=tan12n1n2 m=2n1n2  \Rightarrow {\tan ^{ - 1}}m = {\tan ^{ - 1}}\dfrac{{2n}}{{1 - {n^2}}} \\\ \Rightarrow m = \dfrac{{2n}}{{1 - {n^2}}} \\\

Substituting this value of mm in the equation m=4nm = 4n, we get

2n1n2=4n \Rightarrow \dfrac{{2n}}{{1 - {n^2}}} = 4n

Dividing the above equation by 2n2n on each of the sides, we get

11n2=2 \Rightarrow \dfrac{1}{{1 - {n^2}}} = 2

Cross-multiplying the above equation, we get

1=2(1n2) 1=22n2  \Rightarrow 1 = 2\left( {1 - {n^2}} \right) \\\ \Rightarrow 1 = 2 - 2{n^2} \\\

Subtracting the above equation by 2 on both sides, we get

12=22n22 1=2n2  \Rightarrow 1 - 2 = 2 - 2{n^2} - 2 \\\ \Rightarrow - 1 = - 2{n^2} \\\

Dividing the above equation by 2 - 2 on each of the sides, we get

12=2n22 12=n2 n2=12  \Rightarrow \dfrac{{ - 1}}{{ - 2}} = \dfrac{{ - 2{n^2}}}{{ - 2}} \\\ \Rightarrow \dfrac{1}{2} = {n^2} \\\ \Rightarrow {n^2} = \dfrac{1}{2} \\\

Taking square root in the above equation on both sides, we get

n=±12 \Rightarrow n = \pm \dfrac{1}{{\sqrt 2 }}

Since we know that the slope of a line can never be negative, so the value 12 - \dfrac{1}{{\sqrt 2 }} is discarded.

Substituting the positive value of nn in the given equation m=4nm = 4n, we get

m=42m = \dfrac{4}{{\sqrt 2 }}

Now we will find the product mnmn from these values of mm and nn.

mn=42×12 mn=42 mn=2  mn = \dfrac{4}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow mn = \dfrac{4}{2} \\\ \Rightarrow mn = 2 \\\

Thus, the product mnmn equals 2.