Question

4cos^260°+3sec^230°-cot^245° upon cos^260°+sine^260°

Answer 1

hii


here is ur solution



$\frac{4 \times \frac{1}{4 } + 3 \times \frac{4}{3} - 1 }{ \frac{1}{4} + \frac{3}{4} }$


$\frac{ \frac{4}{4} + \frac{12}{3} - 1 }{ \frac{4}{4} }$


$\frac{1 + \frac{12}{3} - 1}{1}$


$\frac{1 + 4 - 1}{1}$


$\frac{5 - 1}{1}$


$\frac{4}{1} = 4$



hope it helps you