Question

|(3-sec^2x)max -(4+tan^2y)min|

Answer 1

Hello Friend..❤️❤️

The answer of u r question is..✌️✌️


$(3 - {sec}^{2} xmax - (4 + {tan}^{2} ymin$
So,

The answer is..⬇️⬇️

$(1secx)max - (2 \tan \: y)min$

Thank you..⭐️⭐️⭐️

Answer 2

Hey mate.........
here's Ur answer..........

(3- sec ^2x) max - (4+tan ^2y) min.........

=( 1secx )max - ( 2tan y )min

Hope it Helps❤❤