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(C) The given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we

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$\pm \frac{1}{5}$

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(C) The given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = 5 \Rightarrow h = \frac{5}{2}$,and $b = 3$. The angle $	heta$ between the pair of straight lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(\frac{5}{2})^2 - (2)(3)}}{2 + 3} ight| = \left| \frac{2\sqrt{\frac{25}{4} - 6}}{5} ight| = \left| \frac{2\sqrt{\frac{1}{4}}}{5} ight| = \left| \frac{2 	imes \frac{1}{2}}{5} ight| = \frac{1}{5}$. Since the angle is $	an^{-1}(k)$,we have $	an 	heta = k$. Thus,$k = \pm \frac{1}{5}$.

Find the value of $k$,if the angle between the straight lines represented by $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$ is $\tan^{-1}(k)$.

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