Question

Express ${\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right), - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}$ in the simplest form.

Answer 1

To solve this question, we will use some basic trigonometric identities such as, $\cos 2x = {\cos ^2}x - {\sin ^2}x$ and $\sin 2x = 2\sin x\cos x$

Complete step-by-step answer :
We know that,
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
Replaces $x$ by $\dfrac{x}{2}$,
$\Rightarrow \cos 2\dfrac{x}{2} = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$
$\Rightarrow \cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$ ……… (i)
Similarly, we know that
$\sin 2x = 2\sin x\cos x$
Replace $x$ by $\dfrac{x}{2}$,
$\Rightarrow \sin 2\dfrac{x}{2} = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
$\Rightarrow \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ ……… (ii)
We also know that,
${\sin ^2}x + {\cos ^2}x = 1$
Replace $x$ by $\dfrac{x}{2}$,
${\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} = 1$ ………. (iii)
Now, we have
${\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right)$
Putting the value of sin x, cos x and 1 from equation (i), (ii) and (iii), we will get
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} \right)$
Using the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, we can write ${\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ as ${\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)^2}$
Therefore,
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}^2}}}} \right)$
Now, using the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we can write ${\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$ as $\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)$
Thus,
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}{{{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}^2}}}} \right)$
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}}} \right)$
Now, dividing numerator and denominator both by $\cos \dfrac{x}{2}$, we will get
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}{{\cos \dfrac{x}{2}}}}}{{\dfrac{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}}{{\cos \dfrac{x}{2}}}}}} \right)$
Solving this, we will get
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}}}} \right)$ …..(iv) $\therefore \tan \dfrac{x}{2} = \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}$
As we know that,
$\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right) = \dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}}}$ $\therefore \tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$
Put this in equation (iv),
$\Rightarrow {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)} \right)$
$\Rightarrow \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
Hence, we can say that the simplest form of ${\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 - \sin x}}} \right)$ is $\left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$

Note : For solving such questions, we need to remember the trigonometric properties as these questions can only be solved when we remember the properties and formula. This question can also get solved by putting $\cos x$ as $\sin \left( {\dfrac{\pi }{2} - x} \right)$ and $\sin x$ as $\cos \left( {\dfrac{\pi }{2} - x} \right)$ in the given trigonometric expression. Through this, we will get the answer.