Question

If 7sin2A + 3cos2A = 4 then prove that secA+cosecA : 2/root3 + 2

Answer 1

4 sin2 a+ 3sin2 a + 3cos2 a= 4
4 sin2a + 3(1)= 4
4sin2a = 1
sin2a = 1/4
sin a = 1/2
put this value in the first equation
7(1/4) + 3cos2a = 4
3cos2a=4-7/4
3cos2a = 9/4
cos2a = 9/12
cos a = 2/root3
​sec a + cosec a= 1/ cos a+ 1/ sin a
1/2/root+ 1/1/2
2root3 +2