Solveeit Logo

Question

Question: If the rate of change of sine of angle \(\theta \) is \(K\), then the rate of change of its tangent ...

If the rate of change of sine of angle θ\theta is KK, then the rate of change of its tangent is,
(A) K2{{K}^{2}}
(B) 1K2\dfrac{1}{{{K}^{2}}}
(C) KK
(D) 1K\dfrac{1}{K}

Explanation

Solution

Hint: Rate of change of sine of any angle θ\theta is equal to cosine of that angle θ\theta . Also, rate of change of tangent of θ\theta is equal to square of secant of θ\theta . Use, cosθ=1secθ\cos \theta =\dfrac{1}{\sec \theta } and substitute the value of cosθ\cos \theta , obtained by differentiating sinθ\sin \theta .

Complete step-by-step answer:
The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument value. Derivatives are a fundamental tool of calculus, for example, the derivative of the position of a moving object with respect to time is the object’s velocity, this measures how quickly the position of the object changes when time advances. This derivative is referred to as the instantaneous rate of change. Whenever we are asked to find the rate of change of any function, we just find the derivative of that function or in other words, simply differentiate the function.
Now, we come to the question. It has been given that the rate of change of sinθ\sin \theta =K=K. We know that, derivative of sinθ\sin \theta =cosθ=\cos \theta . Therefore, we get, cosθ=K\cos \theta =K………………………………..(i).
Also, we know that derivative of tanθ=sec2θ\tan \theta ={{\sec }^{2}}\theta . Now,secθ=1cosθ\sec \theta =\dfrac{1}{\cos \theta }, therefore, substituting the value of cosθ\cos \theta from equation (i), we get rate of change of tanθ=1cos2θ=1K2\tan \theta =\dfrac{1}{{{\cos }^{2}}\theta }=\dfrac{1}{{{K}^{2}}}.
Hence, option (b) is the correct answer.

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 sinθ\sin \theta and tanθ\tan \theta with respect to angle θ\theta .