Question

The value of ${\operatorname{sech} ^{ - 1}}(\sin \theta )$ is
A) $\log \left( {\tan \dfrac{\theta }{2}} \right)$
B) $\log \left( {\sin \dfrac{\theta }{2}} \right)$
C) $\log \left( {\cos \dfrac{\theta }{2}} \right)$
D) $\log \left( {\cot \dfrac{\theta }{2}} \right)$

Answer 1

Hint: According to given in the question we have to find the value of the given trigonometric expression ${\operatorname{sech} ^{ - 1}}(\sin \theta )$ so, to solve the given trigonometric expression first of all we have to let the expression $y$ now, to solve the obtained expression in form of y we have to convert the trigonometric terms ${\operatorname{sech} ^{ - 1}}(\sin \theta )$in the form of ${\cosh ^{ - 1}}(\cos ec\theta )$ with the help of the formula as given below:

Formula used:
$\sec \theta = \dfrac{1}{{\cos \theta }}..................(1)$
$\sin \theta = \dfrac{1}{{\cos ec\theta }}...............(2)$
Now, after obtaining the trigonometric expression in the form of ${\cosh ^{ - 1}}(\cos ec\theta )$ and to solve the obtain expression we have to use the formula as given below:
${\cosh ^{ - 1}}x = \log \left| {x + \sqrt {{x^2} - 1} } \right|....................(3)$
After applying the formula (3) we will obtain the expression in form of $\cos e{c^2}\theta $ to which we have to convert in the form of $\cot \theta $with the help of the formula given below:
$\cos e{c^2}\theta - 1 = {\cot ^2}\theta ................(4)$
On solving the expression we have get the expression in the form of $1 + \cos \theta $to which we have to covert in the form of $\cos \dfrac{\theta }{2}$with the help of the formula given below:
$1 + \cos \theta = 2\cos e{c^2}\dfrac{\theta }{2}.................(5)$
$\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}................(6)$

Complete step by step answer:
Step 1: To find the value of the given trigonometric expression ${\operatorname{sech} ^{ - 1}}(\sin \theta )$ first of all we have to let the expression $y$
$y = {\operatorname{sech} ^{ - 1}}(\sin \theta )$
And on solving the inverse we can write the expression in the form as given below:
$\operatorname{sech} y = \sin \theta $
Step 2: Now, to solve the obtained expression in form of y we have to convert the trigonometric terms ${\operatorname{sech} ^{ - 1}}(\sin \theta )$ in the form of ${\cosh ^{ - 1}}(\cos ec\theta )$ with the help of the formulas (1) and (2) as mentioned in the solution hint.
$\dfrac{1}{{\sin \theta }} = \dfrac{1}{{\operatorname{sech} y}}$
After cross-multiplication,
$
  \cosh y = \cos ec\theta \\
\Rightarrow y = {\cosh ^{ - 1}}(\cos ec\theta ) \\
 $
Step 3: Now, to solve the obtained expression just above we have to use the formula (3) as mentioned in the solution hint.
$\Rightarrow y = \log \left| {\cos ec\theta + \sqrt {\cos e{c^2}\theta - 1} } \right|$
Step 4: Now, to solve the obtained expression just above have to convert $\cos e{c^2}\theta $ in the form of $\cot \theta $ with the help of the formula (4) as mentioned in the solution hint.
$\Rightarrow y = \log \left| {\cos ec\theta + \cot \theta } \right|$
On solving the obtained expression,
$
\Rightarrow y = \log \left| {\dfrac{1}{{\sin \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}} \right| \\
\Rightarrow y = \log \left| {\dfrac{{1 + \cos \theta }}{{\sin \theta }}} \right| \\
 $
Step 5: Now, we have to convert $1 + \cos \theta $ in the form of $\cos \dfrac{\theta }{2}$with the help of the formulas (5) and (6) as mentioned in the solution hint.
$y = \log \left| {\dfrac{{2{{\cos }^2}\dfrac{\theta }{2}}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right|$
Eliminating $\cos \dfrac{\theta }{2}$ from numerator and denominator,
$
 \Rightarrow y = \log \left| {\dfrac{{\cos \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}}}} \right| \\
\Rightarrow y = \log \left| {\cot \dfrac{\theta }{2}} \right| \\
\Rightarrow y = \log \cot \dfrac{\theta }{2} \\
 $

Hence with the help of the formulas as mentioned in the solution hint we have obtain the value of trigonometric expression ${\operatorname{sech} ^{ - 1}}(\sin \theta ) = \log \cot \dfrac{\theta }{2}$.

Note:
To make the trigonometric expression in easy form it is necessary have to convert the trigonometric terms ${\operatorname{sech} ^{ - 1}}(\sin \theta )$in the form of ${\cosh ^{ - 1}}(\cos ec\theta )$
It is necessary to let the given trigonometric expression be some variables as x, y, or z.