Question

Prove that ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + \cos x} + \sqrt {1 - \cos x} }}{{\sqrt {1 + \cos x} - \sqrt {1 - \cos x} }}} \right) = \dfrac{\pi }{4} - \dfrac{x}{2}$ if $\pi < x < \dfrac{{3\pi }}{2}$.

Answer 1

Hint : We will first use the formulas $1 + \cos 2a = 2{\cos ^2}a$and $1 - \cos 2a = 2{\sin ^2}a$to solve the terms inside the bracket. Then, we will have terms like $\sqrt {{x^2}}$ which we know, $\sqrt {{x^2}} = |x|$. After substituting, we are given the range for $x$, we will then find the range for $\dfrac{x}{2}$as we will consider $x = 2a$ which will imply $a = \dfrac{x}{2}$. After finding the range for $\dfrac{x}{2}$, we will check whether the term in the bracket is positive or negative in that range and then we will consider the signs. If the term is negative then the modulus of that term will have negative sign and if the term is positive then the modulus of that term will have positive sign. After that, we will divide numerator and denominator of the terms in bracket by $\cos \dfrac{x}{2}$and then use the formulas $\tan \left( {\dfrac{\pi }{4} \pm A} \right) = \dfrac{{1 \pm \tan A}}{{1 \mp \tan A}}$. At last, we know, ${\tan ^{ - 1}}(\tan \theta ) = \theta$ if $- \dfrac{\pi }{2} < \theta < \dfrac{\pi }{2}$. We will then consider the range for the angle we will get and solve accordingly.

Complete step by step solution:
We have to prove ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + \cos x} + \sqrt {1 - \cos x} }}{{\sqrt {1 + \cos x} - \sqrt {1 - \cos x} }}} \right) = \dfrac{\pi }{4} - \dfrac{x}{2}$ if $\pi < x < \dfrac{{3\pi }}{2}$. ----(1)
Considering Left Hand Side,
${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {1 + \cos x} + \sqrt {1 - \cos x} }}{{\sqrt {1 + \cos x} - \sqrt {1 - \cos x} }}} \right)$
Taking $2a = x$, we have $a = \dfrac{x}{2}$
Now, Using the formulas $1 + \cos 2a = 2{\cos ^2}a$ and $1 - \cos 2a = 2{\sin ^2}a$ inside the brackets, Left Hand Side becomes
${\tan ^{ - 1}}\left( {\dfrac{{\sqrt {2{{\cos }^2}\dfrac{x}{2}} + \sqrt {2{{\sin }^2}\dfrac{x}{2}} }}{{\sqrt {2{{\cos }^2}\dfrac{x}{2}} - \sqrt {2{{\sin }^2}\dfrac{x}{2}} }}} \right)$
As we know$\sqrt {xy} = \sqrt x \sqrt y$, Left Hand Side becomes
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 \times \sqrt {{{\cos }^2}\dfrac{x}{2}} + \sqrt 2 \times \sqrt {{{\sin }^2}\dfrac{x}{2}} }}{{\sqrt 2 \times \sqrt {{{\cos }^2}\dfrac{x}{2}} - \sqrt 2 \times \sqrt {{{\sin }^2}\dfrac{x}{2}} }}} \right)$
Now, using $\sqrt {{x^2}} = |x|$ in the above expression
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 \times |\cos \dfrac{x}{2}| + \sqrt 2 \times |\sin \dfrac{x}{2}|}}{{\sqrt 2 \times |\cos \dfrac{x}{2}| - \sqrt 2 \times |\sin \dfrac{x}{2}|}}} \right)$
Using Distributive Property i.e. $a(b \pm c) = ab \pm ac$
$= {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 2 (|\cos \dfrac{x}{2}| + |\sin \dfrac{x}{2}|)}}{{\sqrt 2 (|\cos \dfrac{x}{2}| - |\sin \dfrac{x}{2}|)}}} \right)$
Dividing Numerator and denominator inside the bracket by $\sqrt 2$
$= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sqrt 2 (|\cos \dfrac{x}{2}| + |\sin \dfrac{x}{2}|)}}{{\sqrt 2 }}}}{{\dfrac{{\sqrt 2 (|\cos \dfrac{x}{2}| - |\sin \dfrac{x}{2}|)}}{{\sqrt 2 }}}}} \right)$
As $\dfrac{{\sqrt 2 }}{{\sqrt 2 }} = 1$, we will be left with
$= {\tan ^{ - 1}}\left( {\dfrac{{|\cos \dfrac{x}{2}| + |\sin \dfrac{x}{2}|}}{{|\cos \dfrac{x}{2}| - |\sin \dfrac{x}{2}|}}} \right)$ -----(2)
Now, we are given $\pi < x < \dfrac{{3\pi }}{2}$, i.e. $x$lies in third quadrant
Now, dividing by $2$, $\dfrac{\pi }{2} < \dfrac{x}{2} < \dfrac{1}{2} \times \dfrac{{3\pi }}{2}$
$\dfrac{\pi }{2} < \dfrac{x}{2} < \dfrac{{3\pi }}{4}$, i.e. $\dfrac{x}{2}$ lies in Second Quadrant.
We know, $\sin x$is positive if $x$lies in the second quadrant and $\cos x$is negative if $x$lies in the Second Quadrant.
Now, since $\dfrac{x}{2}$lies in second quadrant,
So, $|\sin \dfrac{x}{2}| = \sin \dfrac{x}{2}$ and $|\cos \dfrac{x}{2}| = - \cos \dfrac{x}{2}$ ----(3)
Using (3) in (2), we get Left Hand Side
$= {\tan ^{ - 1}}\left( {\dfrac{{ - \cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}{{ - \cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}} \right)$
Now, Dividing the numerator and denominator inside the bracket by $\cos \dfrac{x}{2}$
$= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{ - \cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}{{\dfrac{{ - \cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}} \right)$
Now, we will split the denominators of both the numerator and denominator.
$= {\tan ^{ - 1}}\left( {\dfrac{{ - \dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}{{ - \dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} - \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}} \right)$
We very well know, $\left( {\dfrac{{\sin x}}{{\cos x}} = \tan x} \right)$. So, using this we get
$= {\tan ^{ - 1}}\left( {\dfrac{{ - 1 + \tan \dfrac{x}{2}}}{{ - 1 - \tan \dfrac{x}{2}}}} \right)$
Now, Taking negative sign common from the numerator as well as the denominator
$= {\tan ^{ - 1}}\left( {\dfrac{{ - \left( {1 - \tan \dfrac{x}{2}} \right)}}{{ - \left( {1 + \tan \dfrac{x}{2}} \right)}}} \right)$
Cancelling the negative sign from numerator and denominator, we are left with
$= {\tan ^{ - 1}}\left( {\dfrac{{\left( {1 - \tan \dfrac{x}{2}} \right)}}{{\left( {1 + \tan \dfrac{x}{2}} \right)}}} \right)$
We know, $\left( {\tan \left( {\dfrac{\pi }{4} \pm A} \right) = \dfrac{{1 \pm \tan A}}{{1 \mp \tan A}}} \right)$. So, now using this identity , we have
$= {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right)$
We are given, $\pi < x < \dfrac{{3\pi }}{2}$
Now, dividing by $2$, we get $\dfrac{\pi }{2} < \dfrac{x}{2} < \dfrac{1}{2} \times \dfrac{{3\pi }}{2} \Rightarrow \dfrac{\pi }{2} < \dfrac{x}{2} < \dfrac{{3\pi }}{4}$
Multiplying with $( - 1)$, we get $- \dfrac{\pi }{2} > - \dfrac{x}{2} > - \dfrac{{3\pi }}{4} \Rightarrow - \dfrac{{3\pi }}{4} < \- \dfrac{x}{2} < \- \dfrac{\pi }{2}$
(Signs get changed when multiplied with $( - 1)$ )
Adding $\dfrac{\pi }{4}$, we get $\dfrac{\pi }{4} - \dfrac{{3\pi }}{4} < \dfrac{\pi }{4} - \dfrac{x}{2} < \dfrac{\pi }{4} - \dfrac{\pi }{2}$
Taking LCM on the left and the right side.
$\Rightarrow \dfrac{{\pi - 3\pi }}{4} < \dfrac{\pi }{4} - \dfrac{x}{2} < \dfrac{{\pi - 2\pi }}{4}$
$\Rightarrow - \dfrac{{2\pi }}{4} < \dfrac{\pi }{4} - \dfrac{x}{2} < \- \dfrac{\pi }{4}$
Dividing the numerator and denominator by $2$ on the left side and Comparing the terms on the right side.
$\Rightarrow - \dfrac{\pi }{2} < \dfrac{\pi }{4} - \dfrac{x}{2} < \- \dfrac{\pi }{4} < \dfrac{\pi }{2}$
$\Rightarrow - \dfrac{\pi }{2} < \dfrac{\pi }{4} - \dfrac{x}{2} < \dfrac{\pi }{2}$ ----(4)
We know, ${\tan ^{ - 1}}(\tan \theta ) = \theta$ if $- \dfrac{\pi }{2} < \theta < \dfrac{\pi }{2}$.
From (4), we have $- \dfrac{\pi }{2} < \dfrac{\pi }{4} - \dfrac{x}{2} < \dfrac{\pi }{2}$
So, ${\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right) = \dfrac{\pi }{4} - \dfrac{x}{2}$
Hence, Proved.

Note : We usually ignore the conditions on $x$ given in the question and solve without considering the conditions. Also, when we use ${\tan ^{ - 1}}(\tan \theta ) = \theta$, we ignore the conditions on $\theta$. We have to remember which functions are positive or negative in which quadrant. And while we are changing the ranges for different angles when we are given a range for a particular angle, we usually forget to change the greater than or less than sign when we multiply with some negative number. Also, all the trigonometric identities are to be remembered thoroughly.