# Trig. Ratios

• May 7th 2014, 06:47 PM
eleventhhour
Trig. Ratios
9) Show that tan30 + 1/tan30 = 1/sin30cos30

So, for this I just subbed in the appropriate trig ratios and solved it to get 2.3, which is the same for both. However, the answer in the book has it as 4√3/3. I know this equals the same thing, but I'm not sure how to get it in that format...Could someone show and explain the steps? Thanks! (:
• May 7th 2014, 06:50 PM
romsek
Re: Trig. Ratios
Just write it as $\dfrac {\sin(30)}{\cos(30)} + \dfrac {\cos(30)}{\sin(30)}$ and simplify
• May 7th 2014, 06:56 PM
Prove It
Re: Trig. Ratios
What Romsek says is I think the easiest way, but you could also evaluate each of the ratios directly...

\displaystyle \begin{align*} LHS &= \tan{ \left( 30^{\circ} \right) } + \frac{1}{\tan{ \left( 30^{\circ} \right) } } \\ &= \frac{ 1}{\sqrt{3}} + \frac{1}{\frac{1}{\sqrt{3}}} \\ &= \frac{\sqrt{3}}{3} + \sqrt{3} \\ &= \frac{4\sqrt{3}}{3} \end{align*}

while

\displaystyle \begin{align*} RHS &= \frac{1}{\sin{ \left( 30^{\circ} \right) } \cos{ \left( 30^{\circ} \right) } } \\ &= \frac{1}{\frac{1}{2} \cdot \frac{\sqrt{3}}{2} } \\ &= \frac{1}{\frac{\sqrt{3}}{4}} \\ &= \frac{4}{\sqrt{3}} \\ &= \frac{4\sqrt{3}}{3} \\ &= LHS \end{align*}
• May 8th 2014, 04:40 PM
eleventhhour
Re: Trig. Ratios
How did you get tan30 = 1/√3? Wouldn't it be √3/3?

And where did the 4 come from in the last part of LHS?
• May 8th 2014, 04:44 PM
romsek
Re: Trig. Ratios
Quote:

Originally Posted by eleventhhour
How did you get tan30 = 1/√3? Wouldn't it be √3/3?

$\dfrac {\frac 1 2}{\frac {\sqrt 3} 2}=\dfrac 1 {\sqrt{3}}=\dfrac {\sqrt{3}} 3$