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Question: If \[{{\sec }^{-1}}\left( \dfrac{1+x}{1-y} \right)=a\] then \[\dfrac{dy}{dx}\] is equal to A. \[\d...

If sec⁡−1(1+x1−y)=a{{\sec }^{-1}}\left( \dfrac{1+x}{1-y} \right)=a then dydx\dfrac{dy}{dx} is equal to
A. y−1x+1\dfrac{y-1}{x+1}
B. y+1x−1\dfrac{y+1}{x-1}
C. x−1y−1\dfrac{x-1}{y-1}
D. x−1y+1\dfrac{x-1}{y+1}

Explanation

Solution

Hint : In this particular problem in first step we have to take sec⁡−1{{\sec }^{-1}} on right hand side to become sec⁡\sec
Then we have to multiply 1−y1-y on both to make the problems easier. Then we have to take derivatives on both sides then apply the derivation rule and solve further then find the value of dydx\dfrac{dy}{dx} .

Complete step-by-step answer :
According to the question it is given that sec⁡−1(1+x1−y)=a{{\sec }^{-1}}\left( \dfrac{1+x}{1-y} \right)=a
So, here we have to find dydx\dfrac{dy}{dx}
First of all we need to take sec⁡−1{{\sec }^{-1}} on right hand side to become sec⁡\sec so that equation becomes
(1+x1−y)=sec⁡(a)\left( \dfrac{1+x}{1-y} \right)=\sec (a)
In this if you take direct then you have to apply uv\dfrac{u}{v} then it becomes little complicated and time consuming to avoid that before taking derivative we need to multiply 1−y1-y on both to become the problems easier we get
(1−y)(1+x1−y)=(1−y)sec⁡(a)\left( 1-y \right)\left( \dfrac{1+x}{1-y} \right)=\left( 1-y \right)\sec (a)
On LHS if you see then 1−y1-y get cancelled and the above equation becomes
1+x=(1−y)sec⁡(a)1+x=\left( 1-y \right)\sec (a)
Take the derivative with respect to x on both side
d(1+x)dx=d((1−y)sec⁡(a))dx\dfrac{d\left( 1+x \right)}{dx}=\dfrac{d\left( \left( 1-y \right)\sec (a) \right)}{dx}
As we know that sec⁡(a)\sec (a) take outside of the derivative on RHS
d(1+x)dx=sec⁡(a)d(1−y)dx\dfrac{d\left( 1+x \right)}{dx}=\sec (a)\dfrac{d\left( 1-y \right)}{dx}
After simplifying further we get:
d(1)dx+d(x)dx=sec⁡(a)[d(1)dx−dydx]\dfrac{d(1)}{dx}+\dfrac{d\left( x \right)}{dx}=\sec (a)\left[ \dfrac{d(1)}{dx}-\dfrac{dy}{dx} \right]
As you can know that d(1)dx=0\dfrac{d(1)}{dx}=0 and d(x)dx=1\dfrac{d\left( x \right)}{dx}=1 substitute this in above equation we get:
0+1=sec⁡(a)[0−dydx]0+1=\sec (a)\left[ 0-\dfrac{dy}{dx} \right]
After simplifying further we get:
1=−sec⁡(a)[dydx]1=-\sec (a)\left[ \dfrac{dy}{dx} \right]
By simplifying further we get:
−1sec⁡(a)=dydx−−(1)\dfrac{-1}{\sec (a)}=\dfrac{dy}{dx}--(1)
If you observe the equation which is given in the question
sec⁡−1(1+x1−y)=a{{\sec }^{-1}}\left( \dfrac{1+x}{1-y} \right)=a
From this equation we get the value of sec⁡(a)\sec (a) that is
sec⁡(a)=(1+x1−y)\sec (a)=\left( \dfrac{1+x}{1-y} \right)
Substitute these above values on equation (1)
−1(1+x1−y)=dydx\dfrac{-1}{\left( \dfrac{1+x}{1-y} \right)}=\dfrac{dy}{dx}
By simplifying further we get:
−(1−y)(1+x)=dydx\dfrac{-(1-y)}{(1+x)}=\dfrac{dy}{dx}
Further solving this we get:
dydx=(y−1)(1+x)\dfrac{dy}{dx}=\dfrac{(y-1)}{(1+x)}
So, the correct answer is “Option A”.

Note : In this particular problem, we can also take direct derivative in first step and solve further and you will get the same answer but the thing is complicated as well as time consuming hence it always advised to do that first we reduce this question and make the problem simpler as much as possible then we take derivative on both sides. Remember all the rules of derivations and simplify the problem step by step.