A) 1
B) \[-1\]
C) \[co{{s}^{2}}\text{ }x\]
D) \[sin{{h}^{2}}\text{ }x\]
Correct Answer: A
Solution :
We have, \[\tan \frac{x}{2}=\tanh =\frac{x}{2}\] \[\Rightarrow \] \[\frac{{{\tan }^{2}}x/2}{1}=\frac{{{\tanh }^{2}}x/2}{1}\] \[\Rightarrow \] \[\frac{1+{{\tan }^{2}}x/2}{1-{{\tan }^{2}}x/2}=\frac{1+{{\tanh }^{2}}x/2}{1-{{\tanh }^{2}}x/2}\] (by using componendo and dividendo rule) \[\Rightarrow \] \[\frac{1}{\cos x}=\cosh x\] \[\Rightarrow \] \[\cos x.\cosh x=1\]
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