Question

Angle of intersection of the curves $r = \sin \theta + cos\theta {\text{ and r = 2sin}}\theta$ is equal to-
$A\dfrac{\pi }{2} \\\ B\dfrac{\pi }{3} \\\ C\dfrac{\pi }{4} \\\ D{\text{ none of these}} \\\$

Answer 1

Here we will proceed by equating both the equations of Angle of intersection of the curves $r = \sin \theta + cos\theta {\text{ and r = 2sin}}\theta$. Then we will simplify the equations using the trigonometric ratios and formulas of the trigonometry table to get the required answer.

Complete step-by-step answer:
As we are given that $r = \sin \theta + cos\theta {\text{ and r = 2sin}}\theta$.
Equating both the equations of angle of intersection of the curves r,
We get-
$2\sin \theta = \sin \theta + \cos \theta$
Or $2\sin \theta - \sin \theta = \cos \theta$
Or $\sin \theta = \cos \theta$
Now dividing both sides by $\cos \theta$ i.e.-
$\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{\cos \theta }}{{\cos \theta }}$
We get-
$\tan \theta = 1$ $\left( {\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta } \right)$
Also we know that $\tan \dfrac{\pi }{4} = 1$
Which implies that-
$\tan \theta = \tan \dfrac{\pi }{4}$
Or $\theta = \dfrac{\pi }{4}$
Therefore, Option C is right.

Note: While solving this question, we must know all the trigonometric ratios of sine, cosine, tangent, cosecant, secant, cotangent as here we used one of these ratios i.e. $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$. Also we must know all the values of the trigonometry table of both of the angles in degrees and angles in radians as here we used one of this formula i.e. $\tan \dfrac{\pi }{4} = 1$.