Question

If 1-cot^2theta/1+cot^2theta=1/2 and theta is an acute angle ,then find the value of theta​

Answer 1

$Given \: \: Question \: \: Is \: \: \\ \\ \frac{1 - \cot {}^{2} (x) }{1 + \cot {}^{2} (x) } = \frac{1}{2} \\ \\ Answer \: \\ \\ \frac{1 - \cot {}^{2} (x) }{ \csc {}^{2} (x) } = \frac{1}{2} \\ \\ \frac{1 - \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) } }{ \csc {}^{2} ( {}^{} x) } = \frac{1}{2} \\ \\ \frac{ \sin {}^{2} (x) - \cos {}^{2} (x) }{ \sin {}^{2} (x) \csc {}^{2} (x) } = \frac{1}{2} \\ \\ \frac{ \sin {}^{2} (x) - \cos {}^{2} (x) }{ \frac{ \sin {}^{2} (x) }{ \sin {}^{2} (x) } } = \frac{1}{2} \\ \\ \sin {}^{2} (x) - \cos {}^{2} (x) = \frac{1}{2} \\ \\ \sin {}^{2} (x) - 1 + \sin {}^{2} (x) = \frac{1}{2} \\ \\ 2\sin {}^{2} (x) = \frac{1}{2} + 1 \\ \\ 2 \sin {}^{2} (x) = \frac{3}{2} \\ \\ \sin {}^{2} (x) = \frac{3}{4} \\ \\ Taking \: square \: \: \: root \: on \: both \: sides \: \\ we \: have \\ \\ \sqrt{ \sin {}^{2} (x) } = \sqrt{ \frac{3}{4} } \\ \\ | \sin(x) | = \frac{ \sqrt{3} }{2} \\ \\ \sin(x) = \frac{ \sqrt{3} }{2} \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \sin(x) = - \frac{ \sqrt{3} }{2} \\ \\ \sin(x) = \sin(60) \: \: \: or \: \: \: \: \sin( x ) = \sin( - 60) \\ \\ x = 60 \: \: \: \: or \: \: \: x = - 60 \\ \\ Note \: \\ \\ 1) \: \: \: \: 1 + \cot {}^{2} (x) = \csc {}^{2} (x) \\ \\ 2) \: \: \: \: \cot {}^{2} (x) = \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) } \\ \\ 3) \: \: \: \: \sin {}^{2} (x) = \frac{1}{ \csc {}^{2} (x) }$