Answer:
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)
![\begin{gathered} (\sin3x-\sin x)/(\cos3x+\cos x)=\tan x \\ (\sin3(30)-\sin30)/(\cos3(30)+\cos30)=0.577 \\ \tan \text{ 30 = 0.577} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b08ouflooerok5tre1uhq0om1ga3680mk4.png)
Therefore:
![(\sin3x-\sin x)/(\cos3x+\cos x)=\tan x](https://img.qammunity.org/2023/formulas/mathematics/college/vj1b53nyx28i8bgld1taygdwwa6dmes3se.png)
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.
![(\sin3x)/(\sin x\cos x)=\text{ 4}\cos x-\sec x](https://img.qammunity.org/2023/formulas/mathematics/college/vzbwajiuqjpl420g0yq7qq0yzn2nbet82u.png)
![\begin{gathered} (\sin 3(30))/(\sin 30\cos 30)=\text{ 2.31} \\ 4\cos 30-\sec 30=\text{ }2.31 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6msh72rothorfp1se24atnhd3t65b7fyyp.png)
Options A to D are correct