(D) We know that $\cosh 2x = \frac{1 + \tanh^2 x}{1 - \tanh^2 x}$.Given $\cosh 2x = 199$,we have:$\frac{1 + \tanh^2 x}{1 - \tanh^2 x} = 199$$1 + \tanh

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(D) We know that $\cosh 2x = \frac{1 + \tanh^2 x}{1 - \tanh^2 x}$.Given $\cosh 2x = 199$,we have:$\frac{1 + \tanh^2 x}{1 - \tanh^2 x} = 199$$1 + \tanh^2 x = 199 - 199 \tanh^2 x$$200 \tanh^2 x = 198$$\tanh^2 x = \frac{198}{200} = \frac{99}{100}$$\tanh x = \sqrt{\frac{99}{100}} = \frac{3 \sqrt{11}}{10}$Since $\coth x = \frac{1}{\tanh x}$,we get:$\coth x = \frac{10}{3 \sqrt{11}}$

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Mix Examples-Trigonometrical Ratios, Functions and Identities

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If $\cosh 2x = 199$,then $\coth x$ equals

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A
$\frac{5}{3 \sqrt{11}}$
B
$\frac{5}{6 \sqrt{11}}$
C
$\frac{7}{3 \sqrt{11}}$
D
$\frac{10}{3 \sqrt{11}}$

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Mix Examples-Trigonometrical Ratios, Functions and Identities

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