The value of 4[1–sin2θ][1+tan2θ]
4
1
2
0
Weknowthat(1−sin2θ=cos2θand 1+tan2θ=sec2θ)
4[1–sin2θ][1+tan2θ] =4 cos2θ×sec2θ=4
Solve: 4[1–sin2θ][1+tan2θ]