The value of θ∈0,π2 and satisfying the equation 1+cos2θsin2θ4sin4θcos2θ1+sin2θ4sin4θcos2θsin2θ1+4sin4θ is
11π24
17π24
5π24
π24
Explanation for the correct option:
Solve the given determinant
Given,
1+cos2θsin2θ4sin4θcos2θ1+sin2θ4sin4θcos2θsin2θ1+4sin4θ=0R3→R3-R2,R2→R2-R11+cos2θsin2θ4sin4θ-1100-11=0C1→C1+C21+cos2θ+sin2θsin2θ4sin4θ-1+1100-1-11=02sin2θ4sin4θ010-1-11=0
Now, if we put 4sin2θ=-2 then the value of the determinant will be 0
∴4sin4θ=-24sin4θ+2=0sin4θ=-12=-sin30°=-sinπ6∴sin4θ=sin-π6[∵-sinθ=sin(-θ)]∴4θ=nπ+(-1)n-π6
∴The θ will be 7π24 and 11π24.
Hence, the correct answer is option (A).