Question

20 POINTS!!!

Answer 1

$\ \ \dfrac{2}{\sqrt{3} \cos (x) + \sin(x)} = \sec\left(\frac{\pi}{6} - x\right)$

Right-hand side

$\text{RHS} = \sec\left(\dfrac{\pi}{6} - x\right)$

Since $\sec x = \frac{1}{\cos x}$, it follows that

$\sec\left(\frac{\pi}{6} - x\right) = \dfrac{1}{\cos\left(\frac{\pi}{6} - x\right) }$

So we can rewrite

$\begin{aligned} \text{RHS} &= \sec\left(\dfrac{\pi}{6} - x\right) \\ &= \dfrac{1}{\cos\left(\frac{\pi}{6} - x\right)} \end{aligned}$

We have a cosine difference identity for the denominator:

$\begin{aligned} \cos(A-B) &= \cos A \cos B + \sin A \sin B \\ \cos\left(\tfrac{\pi}{6} - x\right) &= \cos\left(\tfrac{\pi}{6}\right) \cos (x) + \sin\left(\tfrac{\pi}{6}\right)\sin(x) \end{aligned}$

Since $\sin\left(\tfrac{\pi}{6}\right) = 1/2$ and $\cos\left(\tfrac{\pi}{6}\right) = \sqrt{3}/2$, we have

$\begin{aligned}\cos\left(\tfrac{\pi}{6} - x\right) &= \cos\left(\tfrac{\pi}{6}\right) \cos (x) + \sin\left(\tfrac{\pi}{6}\right)\sin(x) \\ &= \tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x) \end{aligned}$

Using this in the right-hand side

$\begin{aligned} \text{RHS} &= \dfrac{1}{\cos\left(\frac{\pi}{6} + x\right)} \\ &= \frac{1}{\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)} \end{aligned}$

Notice how we have tiny denominators of 2.If we multiply the numerator and denominator of the entire fraction, we will deal with those twos, as 2 will distribute and cancel.

$\begin{aligned} \text{RHS} &= \frac{1}{\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)} \\ &=\frac{2 \cdot(1)}{2 \cdot \left(\tfrac{\sqrt{3}}{2}\cos (x) + \tfrac{1}{2}\sin(x)\right)} \\ &= \frac{2}{\sqrt{3} \cos (x) + \sin(x)} \\ &= \text{LHS} \end{aligned}$

Answer 2

$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$