Question

What is the value of \[{\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right)\]?
A. \[\dfrac{\pi }{4} - \dfrac{x}{2}\]
B.\[\dfrac{\pi }{4} + \dfrac{x}{2}\]
C.\[\dfrac{x}{2}\]
D.\[\dfrac{\pi }{4} - x\]

Answer 1

Hint: To solve the above question first we will use the double angle identities, and then divide the numerator and denominator with \[\cos \dfrac{x}{2}\], and simplifying the expression and finally using trigonometric formula \[\dfrac{{\sin x}}{{\cos x}} = \tan x\], and finally using inverse trigonometric functions \[{\tan ^{ - 1}}\left( {\tan x} \right) = x\] to get the required result.

Formula used: We will use the double-angle formulas such as,
\[\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}\], and \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\],
And we will simplify the \[\sin x\] double angle formula i.e., \[1 + \sin x = {\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)^2}\].
And using the trigonometric formulas \[\dfrac{{\sin x}}{{\cos x}} = \tan x\], and inverse trigonometric formula,
\[{\tan ^{ - 1}}\left( {\tan x} \right) = x\].

Complete Step-by-Step Solution:
Given \[{\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right)\]
Now using the identities \[\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}\],\[1 + \sin x = {\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)^2}\] and we will substitute the values in the expression, we will then get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}^2}}}} \right)\]
Now we will simplify, we will then get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}}{{{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}^2}}}} \right)\]
Now we will eliminate the like terms, then we will get,
\[{\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}}{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}} \right)\]
Now we will divide numerator and denominator with \[\cos \dfrac{x}{2}\], we will get,
\[{\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)}}{{\left( {\dfrac{{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)}}} \right)\]
Now we will simplify by distributing the denominator, we will then get,
\[{\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {\dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} - \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)}}{{\left( {\dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)}}} \right)\]
Now we will trigonometric formula, \[\dfrac{{\sin x}}{{\cos x}} = \tan x\],
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\left( {1 - \tan \dfrac{x}{2}} \right)}}{{\left( {1 + \tan \dfrac{x}{2}} \right)}}} \right)\]
Now we will trigonometric formula i.e., \[\begin{array}{l}\\\dfrac{{1 - \tan \dfrac{x}{2}}}{{1 + \tan \dfrac{x}{2}}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\end{array}\], we will get,
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right)\]
Now again we will inverse trigonometric formula, \[{\tan ^{ - 1}}\left( {\tan x} \right) = x\]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\cos x}}{{1 + \sin x}}} \right) = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]

The correct option is A..

Note: Student are often confused with \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\] and \[\tan \left( {a - b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]. But the correct formula is \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\].