Question

Solve, If $y = \sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} $ then ${\left[ {\dfrac{{dy}}{{dx}}} \right]_{x = 0}}$ is :-
A.$\dfrac{1}{2}$
B.$1$
C.$ - 2$
D. None of the above.

Answer 1

Hint: In order to solve this problem we need to know the formula that $[{(\cos a \pm \sin a)^2} = {\sin ^2}a + {\cos ^2}a \pm 2\sin a\cos a = 1 \pm \sin 2a]$ then differentiate and put x = 2 to get the answer.

Complete Step-by-Step solution:
The given equation is $y = \sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} $ ………(1)
And we know the formula $[{(\cos a \pm \sin a)^2} = {\sin ^2}a + {\cos ^2}a \pm 2\sin a\cos a = 1 \pm \sin 2a]$
Since, ${\sin ^2}a + {\cos ^2}a$=1 and $2\sin a\cos a = \sin 2a$
So we can write equation (1) as :
$
  y = \sqrt {\dfrac{{{{(\cos x - \sin x)}^2}}}{{{{(\cos x + \sin x)}^2}}}} = \dfrac{{(\cos x - \sin x)}}{{(\cos x + \sin x)}} \\
  {\text{On dividing this equation by }}\cos x{\text{ we get:}} \\
   \Rightarrow \dfrac{{1 - \tan x}}{{1 + \tan x}}....................(2) \\
$
And we also know the formula $\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$
$
  \tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}...............(3) \\
  {\text{As we know that, tan}}\dfrac{\pi }{4} = 1 \\
  {\text{so,}}\,\,\dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}} = \dfrac{{1 - \tan x}}{{1 + \tan x}}...............(4) \\
$
From (1),(2), (3) and (4) we can say that :
$\tan \left( {\dfrac{\pi }{4} - x} \right)$=$\dfrac{{1 - \tan x}}{{1 + \tan x}}$
Differentiating $\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} $ means differentiating $\tan \left( {\dfrac{\pi }{4} - x} \right)$ .
So, we can do ${\left[ {\dfrac{{dy}}{{dx}}} \right]_{x = 0}}$ = $\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right) = {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right){\text{ as differentiation of tanx wrt x = se}}{{\text{c}}^2}{\text{x}}{\text{.}}$
On putting x = 0 then we get, ${\sec ^2}\left( {\dfrac{\pi }{4} - 0} \right) = {\sec ^2}\dfrac{\pi }{4} = $2 as $\sec \dfrac{\pi }{4} = \sqrt 2 $.

So, ${\left[ {\dfrac{{dy}}{{dx}}} \right]_{x = 0}}$=2.
Hence the right option is C.

Note: When you have to solve such problems then you have to know the formulas $[{(\cos a \pm \sin a)^2} = {\sin ^2}a + {\cos ^2}a \pm 2\sin a\cos a = 1 \pm \sin 2a]$ and ${\sin ^2}a + {\cos ^2}a$=1 and $2\sin a\cos a = \sin 2a$. Then differentiating according to the general differentiation you will get the right answer.