Question

If \[y = \dfrac{{(\tan (x) + \cot (x))}}{{(\tan (x) - \cot (x))}}\], then \[\dfrac{{dy}}{{dx}} = \]
1) \[2\tan (2x)\sec (2x)\]
2) \[\tan (2x)\sec (2x)\]
3) \[ - \sec (2x)\tan (2x)\]
4) \[ - 2\sec (2x)\tan (2x)\]

Answer 1

Hint: First, simply the expression in terms of a standard trigonometric function and apply the concept of differentiation. While differentiating the function, use the concept of differentiation of essential trigonometric functions. The formulas involved in this question are:
The basic trigonometric formulas like \[\tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}}\] and \[\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}\].
The most common trigonometric identity which is \[{\cos ^2}(x) + {\sin ^2}(x) = 1\]
The formula for \[\cos (2x)\] which is \[\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)\].
Some properties of differentiation as shown below
1) \[\dfrac{{dy}}{{dx}} = \dfrac{{df(x) \times k}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = k\dfrac{{df(x)}}{{dx}}\], where k is a constant
2) \[\dfrac{{dy}}{{dx}} = \dfrac{{df(ax)}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = af'(ax)\], where a is a constant
3) \[\dfrac{{dy}}{{dx}} = \dfrac{{d(\sec (x))}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = \sec (x)\tan (x)\]

Complete answer:
Let’s begin this question by simplifying the expression given
\[ \Rightarrow \dfrac{{(\tan x + \cot x)}}{{(\tan x - \cot x)}}\]
Now, let’s simplify the numerator first,
\[ \Rightarrow (\tan x + \cot x) = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}\]
We know that \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\cot x = \dfrac{{\cos x}}{{\sin x}}\].
Taking LCM, we get,
\[ \Rightarrow \tan x + \cot x = \dfrac{{{{\sin }^2}(x) + {{\cos }^2}(x)}}{{\cos (x) \times \sin (x)}}\]
Clearly, \[{\cos ^2}(x) + {\sin ^2}(x) = 1\], therefore we get,
\[ \Rightarrow \tan x + \cot x = \dfrac{1}{{\cos x\sin x}} - - - - - \left( 1 \right)\]
Now, simplifying the denominator, we have,
\[ \Rightarrow (\tan x - \cot x) = \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\cos x}}{{\sin x}}\]
Now, using \[\cos (2x) = {\cos ^2}(x) - {\sin ^2}(x)\], we get,
\[ \Rightarrow \tan x - \cot x = \dfrac{{{{\sin }^2}(x) - {{\cos }^2}(x)}}{{\cos (x) \times \sin (x)}}\]
Now, we finally get the numerator, as shown below
\[ \Rightarrow \tan x - \cot x = \dfrac{{ - \cos 2x}}{{\cos x\sin x}} - - - - - \left( 2 \right)\]
Now, we have our primary equation as
\[ \Rightarrow y = \dfrac{{(\tan (x) + \cot (x))}}{{(\tan (x) - \cot (x))}}\]
Substituting the value of the numerator and the denominator using the equations \[(i)\] and \[(ii)\], we get,
\[ \Rightarrow y = \dfrac{{\left( {\dfrac{1}{{\cos x\sin x}}} \right)}}{{\left( {\dfrac{{ - \cos 2x}}{{\cos x\sin x}}} \right)}}\]
Now, cancelling the common terms in numerator and denominator, we get,
\[ \Rightarrow y = \left( {\dfrac{{\cos x\sin x}}{{ - \cos 2x\cos x\sin x}}} \right)\]
\[ \Rightarrow y = \left( {\dfrac{1}{{ - \cos 2x}}} \right)\]
Now, we know that secant and cosine are reciprocals of each other. So, we have,
\[ \Rightarrow y = - \sec 2x\]
Now, let us move to the next part of the question, which involves the differentiation of y
Therefore, we have,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d( - \sec (2x))}}{{dx}}\]
Now, using the property of differentiation
\[\dfrac{{dy}}{{dx}} = \dfrac{{df(x) \times k}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = k\dfrac{{df(x)}}{{dx}}\], we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\text{}} - \dfrac{{d(\sec (2x))}}{{dx}}\]
And now, applying another property of differentiation \[\dfrac{{dy}}{{dx}} = \dfrac{{df(ax)}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = af'(ax)\] and \[\dfrac{{dy}}{{dx}} = \dfrac{{d(\sec (x))}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = \sec (x)\tan (x)\]in the above equation we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\text{}} - 2\sec (2x)\tan (2x)\]
Thus, option (4) is the correct answer.

Note:
We should convert the trigonometric functions tangent and cotangent of an angle into sine and cosine of the same angle using the trigonometric formulae: $\tan x = \dfrac{{\sin x}}{{\cos x}}$
and $\cot x = \dfrac{{\cos x}}{{\sin x}}$ as it is relatively easy to work with sine and cosine in place of tangent and cotangent. Chain rule of differentiation $\dfrac{{d\left( {fog\left( x \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$ helps us to differentiate the composite functions layer by layer.