Question

If \[\tan \left( {\theta + i\phi } \right) = \sin \left( {x + i\,y} \right)\], then the value of \[\cot \,hy \sin \,h2\phi \] is
A. \[\tan x\,\,\cot 2\theta \]
B. \[\tan x\,\,\sin 2\theta \]
C. \[cot\,x\,\sin 2\theta \]
D. None of these

Answer 1

Hint: In this question, we need to find the value of \[\coth\,y \sinh\,2\phi \] For that we first simplify the given equation then apply trigonometric formulas and also hyperbolic functions formulas to find the required result.

Formula used:
We have been using the following formulas:
1. \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\]
2. \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\]
3. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
4. \[\tan \theta = \dfrac{1}{{\cot \theta }}\]
5. \[\sin \left( {x + iy} \right) = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
6. \[\sin \left( {i\theta } \right) = i\sinh \theta \]

Complete step-by-step solution:
We are given that \[\tan \left( {\theta + i\phi } \right) = \sin \left( {x + i\,y} \right)...\left( 1 \right)\]
Now we know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]and \[\sin \left( {x + iy} \right) = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
So, our equation (1) becomes:
\[\dfrac{{\sin \left( {\theta + i\phi } \right)}}{{\cos \left( {\theta + i\phi } \right)}} = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
Now we multiply and divide the above equation on the L.H.S by \[2\cos \left( {\theta + i\phi } \right)\]:
\[\dfrac{{\sin \left( {\theta + i\phi } \right)}}{{\cos \left( {\theta + i\phi } \right)}} \times \dfrac{{2\cos \left( {\theta + i\phi } \right)}}{{2\cos \left( {\theta + i\phi } \right)}} = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
Now applying the formula \[2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)\] and \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\], we get
\[\dfrac{{\sin \left( {\theta + i\phi + \theta - i\phi } \right)\, + \,\sin \left( {\theta + i\phi - \theta + i\phi } \right)}}{{\cos \left( {\theta + i\phi + \theta - \phi } \right) + \cos \left( {\theta + i\phi - \theta + i\phi } \right)}} = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
By simplifying we get
\[\dfrac{{\sin 2\theta + i\sinh 2\phi }}{{\cos 2\theta + \cos 2i\phi }} = \sin x\,\cosh y + i\,\cos x\,\sinh y\]
\[\dfrac{{\sin 2\theta + i\sinh 2\phi }}{{\cos 2\theta + i\cosh 2\phi }} = \sin x\,\cosh y + i\cos x\,\sinh y\]
Now divide the above equation into two parts
\[\dfrac{{\sin 2\theta }}{{\cos 2\theta + i\cosh 2\phi }} = \sin x\,\cosh y...\left( 2 \right)\]
\[\dfrac{{\sinh 2\phi }}{{\cos 2\theta + i\cosh 2\phi }} = \cos x\,\sinh y...\left( 3 \right)\]
Now divide equation \[\left( 3 \right)\] by \[\left( 2 \right)\], we get
\[
  \dfrac{{\sin \,h2\phi }}{{\sin 2\theta }} = \dfrac{{\cos x\,\sinh y}}{{\sin x\,\cosh y}} \\
  \dfrac{{\sin \,h\,2\phi }}{{\sin \,2\theta }} = \cot x\tan \,hy \\
  \dfrac{{\sin \,h\,2\phi }}{{\tan \,h\,y}} = \cot x\,\sin 2\theta \\
  \coth\,y\sinh\,2\phi = \cot x\,\sin 2\theta \\
 \]
Therefore, the value of \[\coth\,y \sinh\,2\phi \] is \[\cot x\,\sin 2\theta \].
Hence, option(C) is correct answer

Additional information: Hyperbolic functions are trigonometric functions that are defined for the hyperbola rather than the circle. It is a function of an angle that is described as a relationship between the distances between a point on a hyperbola and the origin and coordinate axes, as hyperbolic sine or hyperbolic cosine, and is commonly stated as a mixture of exponential functions.

Note: Equalities using trigonometric functions are known as trigonometric identities, and they are true for all values of the variables that occur when both sides of the equality are defined. These are geometric identities that involve specific functions of one or more angles. We must apply trigonometric identities to answer the provided types of questions.