Question

Express \[{{\tan }^{-1}}\dfrac{\cos x}{1-\sin x}\], \[\dfrac{-3\pi }{2}< x <\dfrac{\pi }{2}\] in the simplest form.

Answer 1

Hint: In this question, we first need to write the given expression of cosine and sine terms by using trigonometric ratios of multiple angles and the trigonometric identity. Now, further simplify it by using the trigonometric ratios of compound angles formula of tangent. Then by using the properties of the inverse trigonometric functions we can further write in the simplest form.

Complete step-by-step answer:

\[\begin{align}
  & \cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\
 & \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
\end{align}\]
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1+\tan A\tan B}\]
\[{{\tan }^{-1}}\left( \tan \theta \right)=\theta \text{ }\dfrac{-3\pi }{2}<\theta <\dfrac{\pi }{2}\]

Now, let us first consider the given expression in terms of sine and cosine in the question.
\[\Rightarrow \dfrac{\cos x}{1-\sin x}\]
As we already know form the trigonometric ratios of multiple angles that
\[\begin{align}
  & \cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} \\
 & \sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
\end{align}\]
Let us now substitute these values in the above expression to simplify it.
\[\Rightarrow \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{1-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}\]
From the trigonometric identity we know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]
Let us now substitute this trigonometric identity in the above expression
\[\Rightarrow \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}\]
Now, in the above expression denominator can be further written as
\[\Rightarrow \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}}\]
Let us now write the numerator in terms of denominator to simplify it further.
\[\Rightarrow \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\times \left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}}\]
now, on cancelling out the common terms in numerator and denominator we get,
\[\Rightarrow \dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}\]
Now, on taking out the cosine term common in both the numerator and denominator we get,
\[\Rightarrow \dfrac{\cos \dfrac{x}{2}\left( 1+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}{\cos \dfrac{x}{2}\left( 1-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}\]
Now, this can be further simplified and written as
\[\Rightarrow \dfrac{\left( 1+\tan \dfrac{x}{2} \right)}{\left( 1-\tan \dfrac{x}{2} \right)}\]
As we already know that the value \[\tan \dfrac{\pi }{4}=1\] and on substituting this in the above expression we get,
\[\Rightarrow \dfrac{\left( \tan \dfrac{\pi }{4}+\tan \dfrac{x}{2} \right)}{\left( 1-\tan \dfrac{\pi }{4}\cdot \tan \dfrac{x}{2} \right)}\]
As we already know that form the formula of trigonometric ratios of compound angles formula
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1+\tan A\tan B}\]
Now, the above expression can be further written as
\[\Rightarrow \tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)\]
Thus, we get that
\[\Rightarrow \dfrac{\cos x}{1-\sin x}=\tan \left( \dfrac{\pi }{4}+\dfrac{x}{2} \right)\]
Let us now substitute this value of the expression back to get the result.
\[\Rightarrow {{\tan }^{-1}}\dfrac{\cos x}{1-\sin x}\]
\[\Rightarrow {{\tan }^{-1}}\left( tan\left( \dfrac{\pi }{4}+\dfrac{x}{2} \right) \right)\]
As we already know that \[{{\tan }^{-1}}\left( \tan \theta \right)=\theta \]
Now, on further simplification we get,
\[\Rightarrow \dfrac{\pi }{4}+\dfrac{x}{2}\]
Hence, \[{{\tan }^{-1}}\dfrac{\cos x}{1-\sin x}=\dfrac{\pi }{4}+\dfrac{x}{2}\]

Note:
It is important to note that we need to write the cosine and sine terms in terms of their half angles and then further simplify them accordingly so that we can get the common terms in the numerator and the denominator.
It is also important to note that after cancelling the common terms on simplification we further need to convert it to tangent terms because the inverse function is a tangent function.
Instead of using the half angles formula we can also use trigonometric identities and then use the corresponding trigonometric ratios of multiple angles to solve the above question. Both the methods give the same result but it would be a little difficult to solve.