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Question: If \( \cosh (x) = \sec \theta \) , then \( {\tanh ^2}(\dfrac{x}{2}) = \) A. \( {\tan ^2}\dfrac{\t...

If cosh⁡(x)=sec⁡θ\cosh (x) = \sec \theta , then tanh⁡2(x2)={\tanh ^2}(\dfrac{x}{2}) =
A. tan⁡2θ2{\tan ^2}\dfrac{\theta }{2}
B. cot⁡2θ2{\cot ^2}\dfrac{\theta }{2}
C. −tan⁡2θ2- {\tan ^2}\dfrac{\theta }{2}
D. −cot⁡2θ2- {\cot ^2}\dfrac{\theta }{2}

Explanation

Solution

Hint : Hyperbolic functions are defined in terms of the exponential functions; they have similar names to the trigonometric functions. The three main hyperbolic functions are sinhx, coshx and tanhx. Use the definition and identities of these functions to simplify them and you will also require the knowledge of trigonometric identities to find out the correct answer.

Complete step-by-step answer :
We are given that cosh⁡x=sec⁡θ\cosh x = \sec \theta
We know that cosh⁡x=ex+e−x2\cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2}
Use this value in the given equation,
ex+e−x2=sec⁡θ ⇒ex+e−x=2sec⁡θ   \dfrac{{{e^x} + {e^{ - x}}}}{2} = \sec \theta \\\ \Rightarrow {e^x} + {e^{ - x}} = 2\sec \theta \;
We have to find the value of tanh⁡2x2{\tanh ^2}\dfrac{x}{2} .
We know that tanh⁡x=ex−e−xex+e−x\tanh x = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}
That means
tanh⁡x2=ex2−e−x2ex2+e−x2 ⇒tanh⁡2(x2)=(ex2−e−x2ex2+e−x2)2 tanh⁡2x2=(ex2)2+(e−x2)2−2×ex2×e−x2(ex2)2+(e−x2)2+2×ex2×e−x2 tanh⁡2x2=ex+e−x−2ex+e−x+2  \tanh \dfrac{x}{2} = \dfrac{{{e^{\dfrac{x}{2}}} - {e^{\dfrac{{ - x}}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{\dfrac{{ - x}}{2}}}}} \\\ \Rightarrow {\tanh ^2}(\dfrac{x}{2}) = {(\dfrac{{{e^{\dfrac{x}{2}}} - {e^{\dfrac{{ - x}}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{\dfrac{{ - x}}{2}}}}})^2} \\\ {\tanh ^2}\dfrac{x}{2} = \dfrac{{{{({e^{\dfrac{x}{2}}})}^2} + {{({e^{\dfrac{{ - x}}{2}}})}^2} - 2 \times {e^{\dfrac{x}{2}}} \times {e^{\dfrac{{ - x}}{2}}}}}{{{{({e^{\dfrac{x}{2}}})}^2} + {{({e^{\dfrac{{ - x}}{2}}})}^2} + 2 \times {e^{\dfrac{x}{2}}} \times {e^{\dfrac{{ - x}}{2}}}}} \\\ {\tanh ^2}\dfrac{x}{2} = \dfrac{{{e^x} + {e^{ - x}} - 2}}{{{e^x} + {e^{ - x}} + 2}} \\\
Put the value ex+e−x=sec⁡θ{e^x} + {e^{ - x}} = \sec \theta in the above equation,
tanh⁡2x2=2sec⁡θ−22sec⁡θ+2 tanh⁡2x2=sec⁡θ−1sec⁡θ+1 tanh⁡2x2=1cos⁡θ−11cos⁡θ+1 tanh⁡2x2=1−cos⁡θ1+cos⁡θ   {\tanh ^2}\dfrac{x}{2} = \dfrac{{2\sec \theta - 2}}{{2\sec \theta + 2}} \\\ {\tanh ^2}\dfrac{x}{2} = \dfrac{{\sec \theta - 1}}{{\sec \theta + 1}} \\\ {\tanh ^2}\dfrac{x}{2} = \dfrac{{\dfrac{1}{{\cos \theta }} - 1}}{{\dfrac{1}{{\cos \theta }} + 1}} \\\ {\tanh ^2}\dfrac{x}{2} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }} \;
Now, we know that
cos⁡2θ=1−2sin⁡2θ 1−cos⁡2θ=2sin⁡2θ 1−cos⁡θ=2sin⁡2θ2  \cos 2\theta = 1 - 2{\sin ^2}\theta \\\ 1 - \cos 2\theta = 2{\sin ^2}\theta \\\ 1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2} \\\
Also,
cos⁡2θ=2cos⁡2θ−1 1+cos⁡2θ=2cos⁡2θ 1+cos⁡θ=2cos⁡2θ2   \cos 2\theta = 2{\cos ^2}\theta - 1 \\\ 1 + \cos 2\theta = 2{\cos ^2}\theta \\\ 1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2} \;
Putting these values in the obtained equation, we get –
tanh⁡2x2=2sin⁡2θ22cos⁡2θ2=tan⁡2θ2{\tanh ^2}\dfrac{x}{2} = \dfrac{{2{{\sin }^2}\dfrac{\theta }{2}}}{{2{{\cos }^2}\dfrac{\theta }{2}}} = {\tan ^2}\dfrac{\theta }{2}
So, the correct answer is “Option A”.

Note : There are six trigonometric ratios, sine, cosine, tangent, cotangent, cosecant and secant. These six trigonometric ratios are abbreviated as sin, cos, tan, cot, cosec and sec respectively. In hyperbolic functions, the names are the same but their expressions are different. Carefully solve the question, as you may mix up the two functions and get a wrong answer