Question

помогите решить задачу по тригонометрии ​

Answer 1

$\displaystyle\bf\\\frac{2-5Sin^{2} \alpha }{3-5Cos^{2}\alpha } +Sin^{2} \alpha= \frac{2-5Sin^{2} \alpha }{3-5\cdot(1-Sin^{2}\alpha ) } +Sin^{2} \alpha =\\\\\\=\frac{2-5Sin^{2} \alpha }{3-5+5Sin^{2}\alpha } +Sin^{2} \alpha =\frac{2-5Sin^{2} \alpha }{-2+5Sin^{2}\alpha } +Sin^{2} \alpha =\\\\\\=-\frac{2-5Sin^{2} \alpha }{2-5Sin^{2}\alpha } +Sin^{2} \alpha =-1+Sin^{2} \alpha =-(1-Sin^{2} \alpha )=-Cos^{2} \alpha$