The number of values of θ in the interval −π2,π2 satisfying the equation (3)sec2⁡θ=tan4⁡θ+2tan2⁡θ is

tan4⁡θ+2tan2⁡θ=tan2⁡θ+12−1=sec2⁡θ2−1=sec4⁡θ−1putting sec2⁡θ=t we get,(3)t=t2−1 t = 2 is the only solution as t > I Hence, there will be 2 values of θ in given interval.