To prove the trigonometric identity 2/[√3cos(x)+sin(x)] = sec[(π/6)-x], we can follow these steps:
1. Start with the left side of the equation:
2/[√3cos(x)+sin(x)]
2. Multiply the numerator and denominator by the conjugate of the denominator, which is √3cos(x)-sin(x):
2/[√3cos(x)+sin(x)] * [√3cos(x)-sin(x)]/[√3cos(x)-sin(x)]
3. Simplify the expression:
2[√3cos(x)-sin(x)]/[3cos^2(x)-sin^2(x)]
4. Use the trigonometric identity cos^2(x) - sin^2(x) = cos(2x) to rewrite the denominator:
2[√3cos(x)-sin(x)]/[3cos(2x)]
5. Apply the double angle formula cos(2x) = 1 - 2sin^2(x) to further simplify the expression:
2[√3cos(x)-sin(x)]/[3(1 - 2sin^2(x))]
6. Distribute the 3 to the terms inside the brackets:
2[√3cos(x)-sin(x)]/[3 - 6sin^2(x)]
7. Rearrange the terms inside the brackets:
2[√3cos(x)-sin(x)]/[-6sin^2(x) + 3]
8. Factor out a negative sign:
-2[√3cos(x)-sin(x)]/[6sin^2(x) - 3]
9. Divide the numerator and denominator by 3:
-2/3[√3cos(x)-sin(x)]/[2sin^2(x) - 1]
10. Recognize that 2sin^2(x) - 1 is equal to -cos(2x) using the double angle formula:
-2/3[√3cos(x)-sin(x)]/[-cos(2x)]
11. Use the reciprocal identity sec(x) = 1/cos(x) to rewrite the expression as:
-2/3[√3cos(x)-sin(x)]/[-cos(2x)] = -2/3sec(2x)[√3cos(x)-sin(x)]
12. Observe that π/6 - x is the same as 2x:
-2/3sec(2x)[√3cos(x)-sin(x)] = -2/3sec[(π/6)-x][√3cos(x)-sin(x)]
Therefore, we have proven that 2/[√3cos(x)+sin(x)] = sec[(π/6)-x].