Find imaginary part of ${\sin ^{ - 1}}\left( {\cos ec\theta } \right)$
1. $\log \left[ {\cot \left( {\dfrac{\theta }{2}} \right)} \right]$
2. $\dfrac{\pi }{2}$
3. $\left( {\dfrac{1}{2}} \right)\log \left[ {\cot \left( {\dfrac{\theta }{2}} \right)} \right]$
4. None of these
Answer
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Hint: Here in this question, we are given the function ${\sin ^{ - 1}}\left( {\cos ec\theta } \right)$ and we have to find the imaginary part of it. The first step is to let the given function be equal to the general form of a complex number. Take the inverse part to the other sides and apply the inverse trigonometric formula of a complex angle. Then compare the real and imaginary parts of both sides and you’ll get two equations from the equation of the imaginary part calculate the value of $x$ and put it in equation (1) of the real part. Now apply a trigonometric inverse formula to convert the function into a logarithm and solve further.
Formula Used:
Trigonometric Formula of complex angles –
$\sin \left( {x + iy} \right) = \sin x.\cosh y + i\cos x.\sinh y$
Trigonometric formula –
$cose{c^2}\theta - 1 = {\cot ^2}\theta $
Inverse formula –
${\cosh ^{ - 1}}x = \log \left( {x + \sqrt {{x^2} - 1} } \right)$
Complete step by step Solution:
Given that, to find the imaginary part of the function ${\sin ^{ - 1}}\left( {\cos ec\theta } \right)$
Let, ${\sin ^{ - 1}}\left( {cosec\theta } \right) = x + iy$
Therefore, $cosec\theta = \sin \left( {x + iy} \right)$
Using trigonometric formula,
$cosec\theta = \sin x.\cosh y + i\cos x.\sinh y$
Compare the real and imaginary part of both the sides,
$\sin x\cosh y = cosec\theta - - - - - \left( 1 \right)$
$\cos x.\sinh y = 0 - - - - - \left( 2 \right)$
From equation (2),
$\cos x = 0$
Which implies that,$x = \dfrac{\pi }{2}$
Now put the above value in equation (1),
$\sin \left( {\dfrac{\pi }{2}} \right)\cosh y = cosec\theta $
$1\left( {\cosh y} \right) = cosec\theta $
$y = {\cosh ^{ - 1}}\left( {cosec\theta } \right)$
Using trigonometric inverse formula,
$y = \log \left( {cosec\theta + \sqrt {cose{c^2}\theta - 1} } \right)$
Applying the trigonometric identity,
$y = \log \left( {cosec\theta + \cot \theta } \right)$
$y = \log \left( {\cot \dfrac{\theta }{2}} \right)$
Hence, the correct option is 1.
Note: The key concept involved in solving this problem is a good knowledge of complex numbers. Students must remember that complex numbers are made up of two parts: a real number and an imaginary number. Complex numbers serve as the foundation for more complex math, such as algebra. They have many practical applications, particularly in electronics and electromagnetism.
Formula Used:
Trigonometric Formula of complex angles –
$\sin \left( {x + iy} \right) = \sin x.\cosh y + i\cos x.\sinh y$
Trigonometric formula –
$cose{c^2}\theta - 1 = {\cot ^2}\theta $
Inverse formula –
${\cosh ^{ - 1}}x = \log \left( {x + \sqrt {{x^2} - 1} } \right)$
Complete step by step Solution:
Given that, to find the imaginary part of the function ${\sin ^{ - 1}}\left( {\cos ec\theta } \right)$
Let, ${\sin ^{ - 1}}\left( {cosec\theta } \right) = x + iy$
Therefore, $cosec\theta = \sin \left( {x + iy} \right)$
Using trigonometric formula,
$cosec\theta = \sin x.\cosh y + i\cos x.\sinh y$
Compare the real and imaginary part of both the sides,
$\sin x\cosh y = cosec\theta - - - - - \left( 1 \right)$
$\cos x.\sinh y = 0 - - - - - \left( 2 \right)$
From equation (2),
$\cos x = 0$
Which implies that,$x = \dfrac{\pi }{2}$
Now put the above value in equation (1),
$\sin \left( {\dfrac{\pi }{2}} \right)\cosh y = cosec\theta $
$1\left( {\cosh y} \right) = cosec\theta $
$y = {\cosh ^{ - 1}}\left( {cosec\theta } \right)$
Using trigonometric inverse formula,
$y = \log \left( {cosec\theta + \sqrt {cose{c^2}\theta - 1} } \right)$
Applying the trigonometric identity,
$y = \log \left( {cosec\theta + \cot \theta } \right)$
$y = \log \left( {\cot \dfrac{\theta }{2}} \right)$
Hence, the correct option is 1.
Note: The key concept involved in solving this problem is a good knowledge of complex numbers. Students must remember that complex numbers are made up of two parts: a real number and an imaginary number. Complex numbers serve as the foundation for more complex math, such as algebra. They have many practical applications, particularly in electronics and electromagnetism.
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