Question

# If the rate of change of sine of angle $\theta$ is $K$, then the rate of change of its tangent is, (A) ${{K}^{2}}$ (B) $\dfrac{1}{{{K}^{2}}}$ (C) $K$ (D) $\dfrac{1}{K}$

Hint: Rate of change of sine of any angle $\theta$ is equal to cosine of that angle $\theta$. Also, rate of change of tangent of $\theta$ is equal to square of secant of $\theta$. Use, $\cos \theta =\dfrac{1}{\sec \theta }$ and substitute the value of $\cos \theta$, obtained by differentiating $\sin \theta$.
Now, we come to the question. It has been given that the rate of change of $\sin \theta $=K. We know that, derivative of \sin \theta$=\cos \theta$. Therefore, we get, $\cos \theta =K$………………………………..(i).
Also, we know that derivative of $\tan \theta ={{\sec }^{2}}\theta$. Now,$\sec \theta =\dfrac{1}{\cos \theta }$, therefore, substituting the value of $\cos \theta$ from equation (i), we get rate of change of $\tan \theta =\dfrac{1}{{{\cos }^{2}}\theta }=\dfrac{1}{{{K}^{2}}}$.
Note: Rate of change are of two types: (i) instantaneous rate of change and (ii) average rate of change. Here, we have applied an instantaneous rate of change because according to the question we have to find the derivative of $\sin \theta$ and $\tan \theta$ with respect to angle $\theta$.