Question

If $af\left( \tan x \right)+bf\left( \cot x \right)=x,$ then ${{f}^{-1}}\left( \cot x \right)=$
$\left. 1 \right) \dfrac{1}{a-b}$
$\left. 2 \right) \dfrac{{{\sin }^{2}}x}{a+b}$
$\left. 3 \right) \dfrac{{{\sin }^{2}}x}{a-b}$
$\left. 4 \right) \dfrac{{{\sin }^{2}}x}{b-a}$

Answer 1

Hint: We should remember some of the trigonometric identities for solving this problem. We know that the first derivative of the Tangent function is the square of the Secant function. Similarly, the first derivative of the Cotangent function is the negative of the square of the Cosecant function.

Complete step by step solution:
Let us consider the given function, $af\left( \tan x \right)+bf\left( \cot x \right)=x.......\left( 1 \right)$
We are going to find the first derivative of the whole equation given. For that, we need to differentiate the equation once with respect to the variable $x.$
We are going to use the derivatives of trigonometric functions, $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ and $\dfrac{d}{dx}\cot x=-\cos e{{c}^{2}}x.$
Also, we need to remember the facts: when we differentiate $f\left( \tan x \right),$ we first differentiate the function and then differentiate the Tangent function. Similarly, when we differentiate $f\left( \cot x \right),$ we first differentiate the function and then differentiate the Cotangent function.
So, we will get $\dfrac{d}{dx}af\left( \tan x \right)=a\dfrac{d}{dx}f\left( \tan x \right)=a{f}'\left( \tan x \right)\dfrac{d}{dx}\tan x=a{f}'\left( \tan x \right)\left( {{\sec }^{2}}x \right)$
and \[\dfrac{d}{dx}af\left( \cot x \right)=a\dfrac{d}{dx}f\left( \cot x \right)=a{f}'\left( \cot x \right)\dfrac{d}{dx}\cot x=a{f}'\left( \cot x \right)\left( -\cos e{{c}^{2}}x \right).\]
Now we will get $\dfrac{d}{dx}\left( af\left( \tan x \right)+bf\left( \cot x \right) \right)=\dfrac{dx}{dx}.$
Since $\dfrac{dx}{dx}=1,$ we will get $\dfrac{d}{dx}af\left( \tan x \right)+\dfrac{d}{dx}bf\left( \cot x \right)=1.$
We have applied the linearity property of differentiation $\dfrac{d}{dx}\left( au+bv \right)=a\dfrac{du}{dx}+b\dfrac{dv}{dx}$ in the above step.
We can again use the linearity property to get $a\dfrac{d}{dx}f\left( \tan x \right)+b\dfrac{d}{dx}f\left( \cot x \right)=1.$
Now, we can substitute the derivatives of the functions we have found above in the equation obtained, $a{f}'\left( \tan x \right){{\sec }^{2}}x+b{f}'\left( \cot x \right)\left( -\cos e{{c}^{2}}x \right)=1.$
Let us transpose the second summand from the LHS to the RHS, we will get
\[\Rightarrow a{f}'\left( \tan x \right){{\sec }^{2}}x=1-b{f}'\left( \cot x \right)\left( -\cos e{{c}^{2}}x \right)=1+b{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right).\]
Now let us transpose $a{{\sec }^{2}}x$ from the LHS to the RHS,
\[\Rightarrow {f}'\left( \tan x \right)=\dfrac{1+b{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)}{a{{\sec }^{2}}x}.......\left( 2 \right)\]
Now, put $x=\dfrac{\pi }{2}-x$ in the equation $\left( 1 \right).$
We will get $af\left( \tan \left( \dfrac{\pi }{2}-x \right) \right)+bf\left( \cot \left( \dfrac{\pi }{2}-x \right) \right)=\dfrac{\pi }{2}-x.$
Now we will get $af\left( \cot x \right)+bf\left( \tan x \right)=\dfrac{\pi }{2}-x.$
We are going to differentiate the above equation with respect to $x.$
We will get, $\dfrac{d}{dx}\left( af\left( \cot x \right)+bf\left( \tan x \right) \right)=\dfrac{d}{dx}\left( \dfrac{\pi }{2}-x \right).$
Since $\dfrac{d}{dx}\left( \dfrac{\pi }{2}-x \right)=-1,$ we will get $a{f}'\left( \cot x \right)\left( -\cos e{{c}^{2}}x \right)+b{f}'\left( \tan x \right){{\sec }^{2}}x=-1.$
Now let us transpose the first summand from the LHS to the RHS, we will get
\[\Rightarrow b{f}'\left( \tan x \right){{\sec }^{2}}x=-1-a{f}'\left( \cot x \right)\left( -\cos e{{c}^{2}}x \right)=a{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-1.\]
Now let us transpose $b{{\sec }^{2}}x$ from the LHS to the RHS,
\[\Rightarrow {f}'\left( \tan x \right)=\dfrac{a{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-1}{b{{\sec }^{2}}x}.......\left( 3 \right)\]
Now, from the equations $\left( 2 \right)$ and $\left( 3 \right)$ we will get,
\[\Rightarrow {f}'\left( \tan x \right)=\dfrac{1+b{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)}{a{{\sec }^{2}}x}=\dfrac{a{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-1}{b{{\sec }^{2}}x}.\]
Let us cancel ${{\sec }^{2}}x$ from both the fractions,
\[\Rightarrow \dfrac{1+b{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)}{a}=\dfrac{a{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-1}{b}.\]
Transpose $a$ and $b$ to get,
\[\Rightarrow {{a}^{2}}{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-a=b+{{b}^{2}}{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right).\]
Now we will transpose \[b+{{b}^{2}}{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)\] from the RHS to the LHS and $a$ from the LHS and the RHS,
\[\Rightarrow {{a}^{2}}{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)-{{b}^{2}}{f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)=a+b.\]
Now let us take the common factor out,
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right){f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)=a+b.\]
We transpose ${{a}^{2}}-{{b}^{2}}$ from the LHS to the RHS,
\[\Rightarrow {f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)=\dfrac{a+b}{\left( {{a}^{2}}-{{b}^{2}} \right)}.\]
Now we will get,
\[\Rightarrow {f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)=\dfrac{a+b}{\left( a+b \right)\left( a-b \right)}.\]
We cancel $\left( a+b \right)$ from the numerator and the denominator,
\[\Rightarrow {f}'\left( \cot x \right)\left( \cos e{{c}^{2}}x \right)=\dfrac{1}{\left( a-b \right)}.\]
We transpose $\cos e{{c}^{2}}x$ from the LHS to the RHS,
\[\Rightarrow {f}'\left( \cot x \right)=\dfrac{1}{\left( a-b \right)\left( \cos e{{c}^{2}}x \right)}.\]
Since \[\left( {{\sin }^{2}}x \right)=\dfrac{1}{\left( \cos e{{c}^{2}}x \right)},\] we will get \[{f}'\left( \cot x \right)=\dfrac{{{\sin }^{2}}x}{\left( a-b \right)}.\]
Hence \[{f}'\left( \cot x \right)=\dfrac{{{\sin }^{2}}x}{\left( a-b \right)}.\]
So, the correct answer is “Option 3”.

Note: We know that $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x, \cot \left( \dfrac{\pi }{2}-x \right)=\tan x, \sec \left( \dfrac{\pi }{2}-x \right)=\cos ecx$ and $coesc\left( \dfrac{\pi }{2}-x \right)=\sec x.$ Since all the trigonometric functions are positive in the first quadrant, the sign will not change in any of the above identities. Also, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right).$