Question

помогите решить задачу по тригонометрии ​

Answer 1

$\displaystyle\bf\\\frac{2-3Sin^{2} \alpha }{1-3Cos^{2}\alpha } +Cos^{2} \alpha= \frac{2-3Sin^{2} \alpha }{1-3\cdot(1-Sin^{2}\alpha ) } +Cos^{2} \alpha =\\\\\\=\frac{2-3Sin^{2} \alpha }{1-3+3Sin^{2}\alpha } +Cos^{2} \alpha =\frac{2-3Sin^{2} \alpha }{-2+3Sin^{2}\alpha } +Cos^{2} \alpha =\\\\\\=-\frac{2-3Sin^{2} \alpha }{2-3Sin^{2}\alpha } +Cos^{2} \alpha =-1+Cos^{2} \alpha =-(1-Cos^{2} \alpha )=-Sin^{2} \alpha$