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Question: The value of \({\operatorname{sech} ^{ - 1}}(\sin \theta )\) is A) \(\log \left( {\tan \dfrac{\the...

The value of sech⁡−1(sin⁡θ){\operatorname{sech} ^{ - 1}}(\sin \theta ) is
A) log⁡(tan⁡θ2)\log \left( {\tan \dfrac{\theta }{2}} \right)
B) log⁡(sin⁡θ2)\log \left( {\sin \dfrac{\theta }{2}} \right)
C) log⁡(cos⁡θ2)\log \left( {\cos \dfrac{\theta }{2}} \right)
D) log⁡(cot⁡θ2)\log \left( {\cot \dfrac{\theta }{2}} \right)

Explanation

Solution

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

Formula used:
sec⁡θ=1cos⁡θ..................(1)\sec \theta = \dfrac{1}{{\cos \theta }}..................(1)
sin⁡θ=1cos⁡ecθ...............(2)\sin \theta = \dfrac{1}{{\cos ec\theta }}...............(2)
Now, after obtaining the trigonometric expression in the form of cosh⁡−1(cos⁡ecθ){\cosh ^{ - 1}}(\cos ec\theta ) and to solve the obtain expression we have to use the formula as given below:
cosh⁡−1x=log⁡∣x+x2−1∣....................(3){\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⁡ec2θ\cos e{c^2}\theta to which we have to convert in the form of cot⁡θ\cot \theta with the help of the formula given below:
cos⁡ec2θ−1=cot⁡2θ................(4)\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⁡θ1 + \cos \theta to which we have to covert in the form of cos⁡θ2\cos \dfrac{\theta }{2}with the help of the formula given below:
1+cos⁡θ=2cos⁡ec2θ2.................(5)1 + \cos \theta = 2\cos e{c^2}\dfrac{\theta }{2}.................(5)
sin⁡θ=2sin⁡θ2cos⁡θ2................(6)\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 sech⁡−1(sin⁡θ){\operatorname{sech} ^{ - 1}}(\sin \theta ) first of all we have to let the expression yy
y=sech⁡−1(sin⁡θ)y = {\operatorname{sech} ^{ - 1}}(\sin \theta )
And on solving the inverse we can write the expression in the form as given below:
sech⁡y=sin⁡θ\operatorname{sech} y = \sin \theta
Step 2: Now, to solve the obtained expression in form of y we have to convert the trigonometric terms sech⁡−1(sin⁡θ){\operatorname{sech} ^{ - 1}}(\sin \theta ) in the form of cosh⁡−1(cos⁡ecθ){\cosh ^{ - 1}}(\cos ec\theta ) with the help of the formulas (1) and (2) as mentioned in the solution hint.
1sin⁡θ=1sech⁡y\dfrac{1}{{\sin \theta }} = \dfrac{1}{{\operatorname{sech} y}}
After cross-multiplication,
cosh⁡y=cos⁡ecθ ⇒y=cosh⁡−1(cos⁡ecθ)  \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.
⇒y=log⁡∣cos⁡ecθ+cos⁡ec2θ−1∣\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⁡ec2θ\cos e{c^2}\theta in the form of cot⁡θ\cot \theta with the help of the formula (4) as mentioned in the solution hint.
⇒y=log⁡∣cos⁡ecθ+cot⁡θ∣\Rightarrow y = \log \left| {\cos ec\theta + \cot \theta } \right|
On solving the obtained expression,
⇒y=log⁡∣1sin⁡θ+cos⁡θsin⁡θ∣ ⇒y=log⁡∣1+cos⁡θsin⁡θ∣  \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⁡θ1 + \cos \theta in the form of cos⁡θ2\cos \dfrac{\theta }{2}with the help of the formulas (5) and (6) as mentioned in the solution hint.
y=log⁡∣2cos⁡2θ22sin⁡θ2cos⁡θ2∣y = \log \left| {\dfrac{{2{{\cos }^2}\dfrac{\theta }{2}}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right|
Eliminating cos⁡θ2\cos \dfrac{\theta }{2} from numerator and denominator,
⇒y=log⁡∣cos⁡θ2sin⁡θ2∣ ⇒y=log⁡∣cot⁡θ2∣ ⇒y=log⁡cot⁡θ2  \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 sech⁡−1(sin⁡θ)=log⁡cot⁡θ2{\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 sech⁡−1(sin⁡θ){\operatorname{sech} ^{ - 1}}(\sin \theta )in the form of cosh⁡−1(cos⁡ecθ){\cosh ^{ - 1}}(\cos ec\theta )
It is necessary to let the given trigonometric expression be some variables as x, y, or z.