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Question: The angle between the pair of straight lines \({y^2}{\sin ^2}\theta - xy{\sin ^2}\theta + {x^2}\left...

The angle between the pair of straight lines y2sin⁡2θ−xysin⁡2θ+x2(cos⁡2θ−1)=0{y^2}{\sin ^2}\theta - xy{\sin ^2}\theta + {x^2}\left( {{{\cos }^2}\theta - 1} \right) = 0 is:
(A)π4\left( A \right)\dfrac{\pi }{4}
(B)π2\left( B \right)\dfrac{\pi }{2}
(C)π3\left( C \right)\dfrac{\pi }{3}
(D)2π3\left( D \right)\dfrac{{2\pi }}{3}

Explanation

Solution

Hint – In this particular type of question use the concept that the generalized equation of pair of straight lines is given as, (ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0), and the angle between the pair of straight lines is given as,
(θ=tan⁡−1(2h2−aba+b)\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right)) so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Now as we know that the generalized equation of pair of straight lines is given as
ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0.................... (1)
Where, x and y are the variables (i.e. x-axis and y-axis representation).
And a, h and b are the constants.
Now let the angle between the pair of straight lines be (θ\theta ).
Now as we all know that the angle between the pair of straight lines is given as,
⇒θ=tan⁡−1(2h2−aba+b)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right)............. (2)
Now the given equation of the pair of straight lines is
y2sin⁡2θ−xysin⁡2θ+x2(cos⁡2θ−1)=0{y^2}{\sin ^2}\theta - xy{\sin ^2}\theta + {x^2}\left( {{{\cos }^2}\theta - 1} \right) = 0................ (3)
Now compare equation (3) with equation (1) we have,
Therefore, a = cos⁡2θ−1{\cos ^2}\theta - 1, b = sin⁡2θ{\sin ^2}\theta , 2h = −sin⁡2θ- {\sin ^2}\theta
Now divide by 2 throughout in, 2h = −sin⁡2θ- {\sin ^2}\theta we have,
⇒h=−sin⁡2θ2\Rightarrow h = \dfrac{{ - {{\sin }^2}\theta }}{2}
Now substitute the values of a, b and h in equation (2) we have,
⇒θ=tan⁡−1(2(−sin⁡2θ2)2−(cos⁡2θ−1)(sin⁡2θ)sin⁡2θ+cos⁡2θ−1)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{{ - {{\sin }^2}\theta }}{2}} \right)}^2} - \left( {{{\cos }^2}\theta - 1} \right)\left( {{{\sin }^2}\theta } \right)} }}{{{{\sin }^2}\theta + {{\cos }^2}\theta - 1}}} \right)
Now as we know the basic identity of trigonometric (i.e.sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1) so use this value in the above equation we have,
⇒θ=tan⁡−1(2(−sin⁡2θ2)2−(cos⁡2θ−1)(sin⁡2θ)1−1)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{{ - {{\sin }^2}\theta }}{2}} \right)}^2} - \left( {{{\cos }^2}\theta - 1} \right)\left( {{{\sin }^2}\theta } \right)} }}{{1 - 1}}} \right)
Now as we see that the denominator becomes zero. So we have,
⇒θ=tan⁡−1(2(−sin⁡2θ2)2−(cos⁡2θ−1)(sin⁡2θ)0)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{2\sqrt {{{\left( {\dfrac{{ - {{\sin }^2}\theta }}{2}} \right)}^2} - \left( {{{\cos }^2}\theta - 1} \right)\left( {{{\sin }^2}\theta } \right)} }}{0}} \right)
Now as we all know that something divide by zero (0) is infinite so we have,
⇒θ=tan⁡−1(∞)\Rightarrow \theta = {\tan ^{ - 1}}\left( \infty \right)
Now as we all know that tan⁡−1(∞)=90o=π2{\tan ^{ - 1}}\left( \infty \right) = {90^o} = \dfrac{\pi }{2}
Therefore, θ=π2\theta = \dfrac{\pi }{2}
So the angle between the pair of straight lines is 90 degrees.
So this is the required answer.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember the generalized equation of pair of straight lines and the angle between the pair of straight lines which is all stated above, then first compare the given equation with the standard equation and calculate the values of the constant then substitute these values in the formula of the angle and simplify as above we will get the required answer.