Question

show that [sqrt(1-cot^2x)*sqrt(tan^2x)]/(cosx)=tan2x

Answer 1

Show that

$$\frac {\sqrt {1 - \cot^ {2} x} * \sqrt {\tan^ {2} x}}{\cos x} = \tan 2 x$$

Solution:

$$\begin{array}{l} \frac {\sqrt {1 - \cot^ {2} x} * \sqrt {\tan^ {2} x}}{\cos x} = \frac {\sqrt {\left(1 - \left(\frac {\cos x}{\sin x}\right) ^ {2}\right)} * \frac {\sin^ {2} x}{\cos^ {2} x}}{\cos x} = \frac {\sqrt {\left(\frac {\sin^ {2} x - \cos^ {2} x}{\sin^ {2} x}\right)} * \frac {\sin^ {2} x}{\cos^ {2} x}}{\cos x} \\ = \frac {\sqrt {\sin^ {2} x - \cos^ {2} x}}{\cos^ {2} x} \neq \tan 2 x \\ \end{array}$$

For example take $60{}^{\circ}$

$$\tan 60{}^{\circ} = \sqrt {3}$$ $$\cot 60{}^{\circ} = \frac {1}{\sqrt {3}}$$ $$\tan 120{}^{\circ} = - \sqrt {3}$$ $$\cos 60{}^{\circ} = \frac {1}{2}$$ $$\frac {\sqrt {1 - \cot^ {2} x} * \sqrt {\tan^ {2} x}}{\cos x} = \frac {\sqrt {1 - \frac {1}{3}} \sqrt {3}}{2} = \frac {\sqrt {2}}{2}$$ $$\tan 2 x = \tan 120{}^{\circ} = - \sqrt {3}$$ $$\frac {\sqrt {2}}{2} \neq - \sqrt {3}$$