Question

Solve the differentiation $\dfrac{d\left( \sqrt{\dfrac{1-\sin (2x)}{1+\sin (2x)}} \right)}{dx}$
A) ${{\sec }^{2}}(x)$
B) $-{{\sec }^{2}}\left( \dfrac{\pi }{4}-x \right)$
C) ${{\sec }^{2}}\left( \dfrac{\pi }{4}+x \right)$
D) ${{\sec }^{2}}\left( \dfrac{\pi }{4}-x \right)$

Answer 1

In this particular problem, first of all before differentiation we have to simplify the problems by using basic property of trigonometry identity that is ${{\sin }^{2}}(x)+{{\cos }^{2}}(x)=1$as well as half angular formula that is $\sin (2x)=2\sin (x)\cos (x)$and solve further then we have to differentiate with respect to x and get the answer.

Complete step by step solution:
In this particular problem, the expression is given that is
$y=\sqrt{\dfrac{1-\sin (2x)}{1+\sin (2x)}}$
As you can see in this expression we have to simplify this expression by using trigonometry identities as well as half angle formula that is ${{\sin }^{2}}(x)+{{\cos }^{2}}(x)=1$ and $\sin (2x)=2\sin (x)\cos (x)$substitute this formula in above expression we get:
$y=\sqrt{\dfrac{{{\sin }^{2}}(x)+{{\cos }^{2}}(x)-2\sin (x)\cos (x)}{{{\sin }^{2}}(x)+{{\cos }^{2}}(x)+2\sin (x)\cos (x)}}$
To simplify this further we have to basic property of mathematics that is in numerator of the above expression apply ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and in denominator of the above expression apply ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
After applying this we get:
$y=\sqrt{\dfrac{{{(\sin (x)-\cos (x))}^{2}}}{{{(\sin (x)+\cos (x))}^{2}}}}$
It can also be written as
$y=\sqrt{\dfrac{{{(\cos (x)-\sin (x))}^{2}}}{{{(\cos (x)+\sin (x))}^{2}}}}$
After simplifying this we get:
$y=\dfrac{\cos (x)-\sin (x)}{\cos (x)+\sin (x)}$
By divide $\cos (x)$on both sides on numerator and denominator we get:
$y=\dfrac{\dfrac{\cos (x)-\sin (x)}{\cos (x)}}{\dfrac{\cos (x)+\sin (x)}{\cos (x)}}$
By further simplifying we get:
$y=\dfrac{1-\tan (x)}{1+\tan (x)}$
As we know that $\tan \left( \dfrac{\pi }{4} \right)=1$and use this in above expression we get:
$y=\dfrac{\tan \left( \dfrac{\pi }{4} \right)-\tan (x)}{1+\tan (x)\tan \left( \dfrac{\pi }{4} \right)}$
If you observe this above equation you can notice that we can apply the formula of $\tan (A-B)=\dfrac{\tan (A)-\tan (B)}{1+\tan (A)\tan (B)}$
After applying this we get:
$y=\tan \left( \dfrac{\pi }{4}-x \right)$
Now, we have differentiate with respect to x and apply $\dfrac{d\tan (x)}{dx}={{\sec }^{2}}(x)$ in above equation we get:
$\dfrac{dy}{dx}=\left[ {{\sec }^{2}}\left( \dfrac{\pi }{4}-x \right) \right]\times \dfrac{d\left( \dfrac{\pi }{4}-x \right)}{dx}$
After simplifying we get:
$\dfrac{dy}{dx}=-{{\sec }^{2}}\left( \dfrac{\pi }{4}-x \right)$
Therefore, the correct option is option (B).

Note:
In this type of problem always remember the formula of trigonometry as well as the formula of derivative correctly. It is always advisable that first you have to reduce the expression which is given in the question so that the problem will become easier to differentiate and also it can avoid silly mistakes because if you differentiate directly then it becomes lengthy as well as complicated and time consuming. So, the above solution is preferred for such types of problems.