All the options are correct
Explanations:A quick and smart way is to substitute a value for x in each of the options and verify if the right hand side equals the left hand side
Let x = 30
A) (sin x + cos x)² = 1 + sin 2x
(sin 30 + cos 30)² = 1.866
1 + sin 2(30) = 1.866
Therefore (sin x + cos x)² = 1 + sin 2x
B)
[tex]\begin{gathered} \frac{\sin3x-\sin x}{\cos3x+\cos x}=\tan x \\ \frac{\sin3(30)-\sin30}{\cos3(30)+\cos30}=0.577 \\ \tan \text{ 30 = 0.577} \end{gathered}[/tex]Therefore:
[tex]\frac{\sin3x-\sin x}{\cos3x+\cos x}=\tan x[/tex]C) sin 6x = 2 sin3x cos3x
sin 6(30) = 0
2 sin3(30) cos3(30) = 0
Therefore sin 6x = 2 sin3x cos3x
This can also be justified by sin2A = 2sinAcosA
D.
[tex]\frac{\sin3x}{\sin x\cos x}=\text{ 4}\cos x-\sec x[/tex][tex]\begin{gathered} \frac{\sin 3(30)}{\sin 30\cos 30}=\text{ 2.31} \\ 4\cos 30-\sec 30=\text{ }2.31 \end{gathered}[/tex]Options A to D are correct
