Answer
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Hint: Write down the given statement in the differential form and use the differentiation formula of standard trigonometric equations to solve this question. We also need to take help of the general solution of trigonometric equations to get the final result.
Complete step by step solution: We have been given that the rate of change of sine and tangent of the same angle are equal.
Let us assume that the angle is $\theta $.
Thus, from the given statement in the question we can write,
$\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) = \dfrac{d}{{d\theta }}\left( {\tan \theta } \right)$
Using the formula of differentiation of standard trigonometric terms, we get
$\cos \theta = {\sec ^2}\theta $
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$ , thus, the above equation can be written as
$\begin{array}{l}
\cos \theta = \dfrac{1}{{{{\cos }^2}\theta }}\\
\Rightarrow {\cos ^3}\theta = 1
\end{array}$
Taking cube roots on both sides we get
$\cos \theta = 1$
We know that for a particular value in $\left[ {0,2\pi } \right]$ , $\cos 0 = 1$
Since no particular value is required in this question, we can write the right hand side of the above equation as
$\begin{array}{l}
1 = \cos \left( {2n\pi + 0} \right)\;,n \in I\\
\Rightarrow 1 = \cos \left( {2n\pi } \right)\;,n \in I
\end{array}$
We know from general solution of trigonometric equations we get
$\begin{array}{l}
\cos \theta = \cos \left( {2n\pi } \right)\\
\Rightarrow \theta = 2n\pi \;,\;n \in I
\end{array}$
Thus, the third option is the correct option.
Note: Rate of change is a rate which tells us how one quantity changes with respect to another quantity. if we have a function say y = f(x) and one quantity y varies with another quantity x then dy/dx = f’(x). The equations that involve the trigonometric functions of a variable are called trigonometric equations. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation and it is expressed in a generalized form in terms of ‘n’. The general solution of $\cos \theta = \cos \alpha \Rightarrow \theta = 2n\pi \pm \alpha ,n \in I$
Complete step by step solution: We have been given that the rate of change of sine and tangent of the same angle are equal.
Let us assume that the angle is $\theta $.
Thus, from the given statement in the question we can write,
$\dfrac{d}{{d\theta }}\left( {\sin \theta } \right) = \dfrac{d}{{d\theta }}\left( {\tan \theta } \right)$
Using the formula of differentiation of standard trigonometric terms, we get
$\cos \theta = {\sec ^2}\theta $
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$ , thus, the above equation can be written as
$\begin{array}{l}
\cos \theta = \dfrac{1}{{{{\cos }^2}\theta }}\\
\Rightarrow {\cos ^3}\theta = 1
\end{array}$
Taking cube roots on both sides we get
$\cos \theta = 1$
We know that for a particular value in $\left[ {0,2\pi } \right]$ , $\cos 0 = 1$
Since no particular value is required in this question, we can write the right hand side of the above equation as
$\begin{array}{l}
1 = \cos \left( {2n\pi + 0} \right)\;,n \in I\\
\Rightarrow 1 = \cos \left( {2n\pi } \right)\;,n \in I
\end{array}$
We know from general solution of trigonometric equations we get
$\begin{array}{l}
\cos \theta = \cos \left( {2n\pi } \right)\\
\Rightarrow \theta = 2n\pi \;,\;n \in I
\end{array}$
Thus, the third option is the correct option.
Note: Rate of change is a rate which tells us how one quantity changes with respect to another quantity. if we have a function say y = f(x) and one quantity y varies with another quantity x then dy/dx = f’(x). The equations that involve the trigonometric functions of a variable are called trigonometric equations. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation and it is expressed in a generalized form in terms of ‘n’. The general solution of $\cos \theta = \cos \alpha \Rightarrow \theta = 2n\pi \pm \alpha ,n \in I$
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