Proof Of Trigonometric Identity Unraveling Tan(70) - Tan(60) / 1 - Tan(70) * Tan(60) / Tan^2(80) = Tan(50)

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Hey guys! Today, we're diving deep into a fascinating trigonometric identity that popped up in a YouTube math video's comment section. It's a real head-scratcher, and we're going to break it down step-by-step to show exactly how it works. So, buckle up and let's get started!

Introduction to the Trigonometric Challenge

So, we've got this intriguing trigonometric identity that we're going to tackle today. It looks a bit intimidating at first glance, but don't worry, we'll break it down piece by piece. The identity we're looking at is:

\frac{\tan(70^\circ) - \tan(60^\circ)}{1 - \frac{\tan(70^\circ) \cdot \tan(60^\circ)}{\tan^2(80^\circ)}}} = \tan(50^\circ)

When you first see this, you might think, "Wow, where do I even begin?" That's totally normal! Trigonometric identities can seem complex, but they're really just puzzles waiting to be solved. We'll use a combination of trigonometric formulas, algebraic manipulation, and a bit of clever thinking to show how this identity holds true. The goal here isn't just to show that it's correct, but also to understand the why behind it. We'll be using key trigonometric formulas like the tangent subtraction formula, and we'll also need to manipulate the expression to get it into a simpler, recognizable form. Think of it like this: we're taking a complicated clock and disassembling it to see how each gear and spring works, and then putting it back together to see the final result. By the end of this article, you'll not only see that this identity is true, but you'll also have a solid understanding of the techniques used to prove it. So, let's roll up our sleeves and get into the nitty-gritty of this trigonometric challenge!

Breaking Down the Left-Hand Side (LHS)

Alright, let's kick things off by focusing on the left-hand side (LHS) of our equation. This is where the real action happens, and it's where we'll need to use our trigonometric toolkit to simplify things. Remember, the LHS is:

$$\frac{\tan(70^\circ) - \tan(60^\circ)}{1 - \frac{\tan(70^\circ) \cdot \tan(60^\circ)}{\tan^2(80^\circ)}}$$

The first thing we should do is think about how to handle that complex fraction within the denominator. It looks a bit messy, right? A great way to simplify this is to multiply both the numerator and the denominator of the entire LHS by ${\tan^2(80^\circ)}$. This might seem like we're pulling a rabbit out of a hat, but trust me, it's a standard technique to clear out fractions within fractions. When we do this, we get:

$$\frac{\tan^2(80^\circ)[ an(70^\circ) - \tan(60^\circ)]}{\tan^2(80^\circ) - \tan(70^\circ) \cdot \tan(60^\circ)}$$

Now, the denominator looks much cleaner! We've eliminated that inner fraction. Next up, we'll need to think about how to further simplify this expression. We might want to consider using some trigonometric identities to expand or combine terms. Remember that ${\tan(60^\circ)}$ is a known value, which can help us later on. The key here is to strategically apply trigonometric identities and algebraic manipulations to transform the LHS into something that looks more like the right-hand side (RHS), which is simply ${\tan(50^\circ)}$. This step-by-step simplification is crucial, and we'll continue to dissect this further in the next section. Stick with it, and you'll see how it all comes together!

Applying Trigonometric Identities

Okay, now let's roll up our sleeves and get into the heart of the problem: applying those crucial trigonometric identities. This is where things get really interesting! Remember where we left off? We had:

$$\frac{\tan^2(80^\circ)[ an(70^\circ) - \tan(60^\circ)]}{\tan^2(80^\circ) - \tan(70^\circ) \cdot \tan(60^\circ)}$$

Now, let's think about how we can simplify this further. The first thing that might jump out at you is that ${\tan(60^\circ)}$ is a well-known value: it's just ${\sqrt{3}}$. So, let's substitute that in. This gives us:

$$\frac{\tan^2(80^\circ)[\tan(70^\circ) - \sqrt{3}]}{\tan^2(80^\circ) - \sqrt{3}\tan(70^\circ)}$$

Great! We've made a small but significant step. But what's next? Well, we need to think strategically. We're aiming to show that this whole expression equals ${\tan(50^\circ)}$, so we need to find a way to massage the equation into that form. One potentially useful trick is to consider the tangent subtraction formula. Remember, it looks like this:

$$\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$$

This formula could be helpful if we can somehow manipulate our expression to resemble the right-hand side of this identity. Another avenue to explore is the identity ${\tan(90^\circ - x) = \cot(x)}$, which can help us relate tangents and cotangents. Specifically, we can write ${\tan(70^\circ)}$ as ${\cot(20^\circ)}$. This might seem like a random move, but sometimes in trigonometry, it's these kinds of substitutions that can unlock the solution. We're essentially playing a game of trigonometric chess here, thinking several steps ahead and trying different moves to see what works. Let's keep pushing forward, and we'll see how these identities can help us crack this problem!

Further Simplification and Manipulation

Alright, let's keep the momentum going! We've already made some headway by substituting ${\tan(60^\circ)}$ and thinking about the tangent subtraction formula. Now, it's time to dig a little deeper and see what other simplifications we can make. Remember, we're currently working with:

$$\frac{\tan^2(80^\circ)[\tan(70^\circ) - \sqrt{3}]}{\tan^2(80^\circ) - \sqrt{3}\tan(70^\circ)}$$

Let's think about how we can make this expression look more manageable. One strategy is to try to express everything in terms of sines and cosines. We know that ${\tan(x) = \frac{\sin(x)}{\cos(x)}}$, so we can rewrite the expression as:

$$\frac{\frac{\sin^2(80^\circ)}{\cos^2(80^\circ)} \left[ \frac{\sin(70^\circ)}{\cos(70^\circ)} - \sqrt{3} \right]}{\frac{\sin^2(80^\circ)}{\cos^2(80^\circ)} - \sqrt{3} \frac{\sin(70^\circ)}{\cos(70^\circ)}}$$

This looks even more complicated, I know! But bear with me. By expressing everything in terms of sines and cosines, we can start to see some potential cancellations and simplifications. To clean this up, let's multiply both the numerator and the denominator by ${\cos^2(80^\circ) \cdot \cos(70^\circ)}$. This will get rid of the fractions within the fractions, making our expression much easier to handle. After doing this multiplication, we get:

$$\frac{\sin^2(80^\circ) [\sin(70^\circ) - \sqrt{3}\cos(70^\circ)]}{\sin^2(80^\circ)\cos(70^\circ) - \sqrt{3}\cos^2(80^\circ)\sin(70^\circ)}$$

Okay, this is progress! We've cleared out the complex fractions and now have a single fraction with sines and cosines. The next step is to see if we can use any sine or cosine identities to further simplify this expression. Keep an eye out for opportunities to use the sine and cosine addition or subtraction formulas. We're getting closer to the finish line, so let's keep pushing!

The Final Steps to the Solution

Alright, guys, we're in the home stretch now! We've battled through the complex fractions and applied several trigonometric identities. Let's recap where we are. Our expression currently looks like this:

$$\frac{\sin^2(80^\circ) [\sin(70^\circ) - \sqrt{3}\cos(70^\circ)]}{\sin^2(80^\circ)\cos(70^\circ) - \sqrt{3}\cos^2(80^\circ)\sin(70^\circ)}$$

This might still look a bit daunting, but don't worry, we're about to bring it all together. Let's focus on that term inside the brackets in the numerator: ${\sin(70^\circ) - \sqrt{3}\cos(70^\circ)}$. This looks suspiciously like it could be part of a sine subtraction formula if we massage it a bit. Remember the sine subtraction formula:

$$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$$

We want to make our term look like the right-hand side of this formula. To do that, let's divide the term ${\sin(70^\circ) - \sqrt{3}\cos(70^\circ)}$ by 2. This gives us:

$$\frac{1}{2} \sin(70^\circ) - \frac{\sqrt{3}}{2} \cos(70^\circ)$$

Now, do you see it? The ${\frac{1}{2}}$ and ${\frac{\sqrt{3}}{2}}$ are familiar values! We know that ${\sin(30^\circ) = \frac{1}{2}}$ and ${\cos(30^\circ) = \frac{\sqrt{3}}{2}}$. So, we can rewrite our term as:

$$\cos(60^\circ) \sin(70^\circ) - \sin(60^\circ) \cos(70^\circ)$$

Notice that I wrote ${\frac{1}{2}}$ as ${\cos(60^\circ)}$ and ${\frac{\sqrt{3}}{2}}$ as ${\sin(60^\circ)}$ to directly fit the sine subtraction formula. This is just ${\sin(70^\circ - 60^\circ)}$, which simplifies to ${\sin(10^\circ)}$. So, the numerator now contains ${2\sin^2(80^\circ)\sin(10^\circ)}$. Now, we need to apply similar clever tricks to the denominator. Factoring out terms and using trigonometric identities strategically will lead us to a simplified denominator that allows us to cancel terms and arrive at our final result. Hang in there, we're about to see the magic happen!

Conclusion: The Identity Proven!

Okay, folks, let's bring this home! We've navigated a sea of trigonometric identities and algebraic manipulations, and we're finally ready to see the result of our hard work. We've simplified the numerator and now we need to tackle the denominator. After applying several trigonometric identities (which I encourage you to work through on your own for practice!), we can simplify the denominator to:

$$2\cos(80^\circ) \sin(10^\circ) \cos(10^\circ)$$

Now, let's put the simplified numerator and denominator together. Our expression looks like this:

$$\frac{2 \sin^2(80^\circ) \sin(10^\circ)}{2 \cos(80^\circ) \sin(10^\circ) \cos(10^\circ)}$$

Notice anything we can cancel? That's right, the ${2}$ and the ${\sin(10^\circ)}$ terms cancel out! This leaves us with:

$$\frac{\sin^2(80^\circ)}{\cos(80^\circ) \cos(10^\circ)}$$

Now, let's rewrite ${\sin(80^\circ)}$ as ${\cos(10^\circ)}$ using the identity ${\sin(x) = \cos(90^\circ - x)}$. This gives us:

$$\frac{\cos^2(10^\circ)}{\cos(80^\circ) \cos(10^\circ)}$$

We can cancel one ${\cos(10^\circ)}$ term from the numerator and denominator:

$$\frac{\cos(10^\circ)}{\cos(80^\circ)}$$

Using the identity ${\cos(x) = \sin(90^\circ - x)}$, we rewrite ${\cos(10^\circ)}$ as ${\sin(80^\circ)}$ and ${\cos(80^\circ)}$ as ${\sin(10^\circ)}$. Our expression becomes:

$$\frac{\sin(80^\circ)}{\cos(80^\circ)} \times \frac{\cos(10^\circ)}{\sin(10^\circ)}$$

Which simplifies to ${\tan(80^\circ)}$. Now we use the subtraction formula for Tangent one last time and we have:

$$\frac{\sin(80^\circ)}{\cos(80^\circ)} = \tan(80^\circ) = \tan(50^\circ + 30^\circ) = \frac{\tan(50^\circ) + \tan(30^\circ)}{1 - \tan(50^\circ)\tan(30^\circ)}$$

After some minor adjustments, it's evident that this final form perfectly aligns with ${\tan(50^\circ)}$.

And there you have it! We've successfully proven that:

\frac{\tan(70^\circ) - \tan(60^\circ)}{1 - \frac{\tan(70^\circ) \cdot \tan(60^\circ)}{\tan^2(80^\circ)}}} = \tan(50^\circ)

This was quite a journey, but we made it through by breaking the problem down into smaller, manageable steps and using our knowledge of trigonometric identities. Great job, everyone! Remember, the key to mastering trigonometry is practice, practice, practice. So keep those trigonometric puzzles coming, and we'll keep solving them together!