Question

Differentiate with respect to \[x,\]
\[{\text{ }}y{\text{ }} = {\text{ }}\left( {tan{\text{ }}x{\text{ }} + {\text{ }}sec{\text{ }}x} \right){\text{ }}\left( {cot{\text{ }}x{\text{ }} + {\text{ }}cosec{\text{ }}x} \right)\]

Answer 1

Hint: For this type of trigonometric problem first we simplify by multiplying the right hand side of a given problem and writing the result as in terms of addition or subtraction. As we know that differentiation of individual terms are easier.
Formulas used: Product rule of differentiation$\dfrac{d}{{dx}}(A.B) = A\dfrac{d}{{dx}}(B) + B\dfrac{d}{{dx}}(A)$, $\tan x = \dfrac{{\sin x}}{{\cos x}},\,\,\sec x = \dfrac{1}{{\cos x}},$ $\,\,\tan x.\cot x = 1$
$\,\,\cot x = \dfrac{{\cos x}}{{\sin x}},\,\,\cos ecx = \dfrac{1}{{\sin x}}$

Complete step by step solution:
Given equation is
\[y = \left( {tan{\text{ }}x + sec{\text{ }}x} \right){\text{ }}\left( {cot{\text{ }}x + cosec{\text{ }}x} \right)\]
Simplifying right hand side of above equation we have,
\[y{\text{ }} = {\text{ }}tan{\text{ }}x{\text{ }}.{\text{ }}cot{\text{ }}x{\text{ }} + {\text{ }}tan{\text{ }}x{\text{ }}cosec{\text{ }}x{\text{ }} + {\text{ }}sec{\text{ }}x{\text{ }}.{\text{ }}cot{\text{ }}x{\text{ }} + {\text{ }}sec{\text{ }}x{\text{ }}.{\text{ }}cosec{\text{ }}x\]
For trigonometric functions we have$\tan x.\cot x = 1$, $\sec x = \dfrac{1}{{\cos x}}and\,\,\cos ecx = \dfrac{1}{{\sin x}}$ . Using these in above we have,
\[y = 1 + \dfrac{{\operatorname{sinx} }}{{\cos x\,}} \times \dfrac{1}{{\sin x}} + \dfrac{1}{{\cos x}} \times \dfrac{{\cos x}}{{\sin x}} + \dfrac{1}{{\sin x}} \times \dfrac{1}{{\cos x}}\]
$ \Rightarrow y = 1 + \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}} + \dfrac{1}{{\sin x}}.\dfrac{1}{{\cos x}}$
Or
$
  y = 1 + \sec x + \cos ecx + \cos ecx.\sec x \\
   \Rightarrow y = (1 + \sec x) + \cos ecx(1 + \sec x) \\
   \Rightarrow y = (1 + \sec x)(1 + \cos ecx) \\
 $
Differentiate above formed equation with respect to x by using product rule.
\[\Rightarrow y = \left( {1 + \cos ec\,x} \right)\,\,\left( {1 + \sec x} \right)\]
$\dfrac{{dy}}{{dx}} = (1 + \sec x)\dfrac{d}{{dx}}(1 + cosecx) + (1 + cosecx)\dfrac{d}{{dx}}(1 + \sec x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because \dfrac{d}{{dx}}\left( {A.B} \right) = A\dfrac{d}{{dx}}(B) + B\dfrac{d}{{dx}}(A)} \right\}$
\[\dfrac{{dy}}{{dx}} = \left( {1 + \sec \,\,x} \right)\left( {cosec\,\,x\,\,\cot \,\,x} \right)\,\, + \left( {1 + \cos ec\,\,x} \right)\,\,\left( {\sec \,\,x\,\,.\,\,\tan \,x} \right)\] $\left\{ {\because \dfrac{d}{{dx}}(\cos ecx) = \cos ecx.\cot x,\,\,\dfrac{d}{{dx}}(\sec x) = \sec x.\tan x} \right\}$
Which is the required derivative of a given function.
Hence, from above we see that derivative of \[{\text{ }}y{\text{ }} = {\text{ }}\left( {tan{\text{ }}x{\text{ }} + {\text{ }}sec{\text{ }}x} \right){\text{ }}\left( {cot{\text{ }}x{\text{ }} + {\text{ }}cosec{\text{ }}x} \right)\] is$\left( {1 + \sec x} \right)\left( {\cos ecx.\cot x} \right) + \left( {1 + \cos ecx} \right)\left( {\sec x.\tan x} \right)$.
So, the correct answer is “Option C”.

Note: In trigonometric functions taking direct derivatives of given terms without changing them to basic t-ratio of trigonometric may lead to different answers as converting given problems in basic trigonometric functions and then differentiating them. But in some cases the answer in both cases will be the same.