Question

If $2 \tan 10^\circ + \tan 50^\circ = 2x$, $\tan 20^\circ + \tan 50^\circ = 2y$, $2 \tan 10^\circ + \tan 70^\circ = 2w$ and $\tan 20^\circ + \tan 70^\circ = 2z$, then which of the following is/are true -

• A. z > w > y > x
• B. w = x + y
• C. 2y = z

Answer 1

Answer: (A), (B), (C)

(1), (2), (3)

Answer 2

Let's analyze each expression and the given options.

The given expressions are:

1. $2 \tan 10^\circ + \tan 50^\circ = 2x$
2. $\tan 20^\circ + \tan 50^\circ = 2y$
3. $2 \tan 10^\circ + \tan 70^\circ = 2w$
4. $\tan 20^\circ + \tan 70^\circ = 2z$

We will simplify $2y$ and $2z$ using the identity $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.

For $2y$: $2y = \tan 20^\circ + \tan 50^\circ = \frac{\sin(20^\circ+50^\circ)}{\cos 20^\circ \cos 50^\circ} = \frac{\sin 70^\circ}{\cos 20^\circ \cos 50^\circ}$ Since $\sin 70^\circ = \cos(90^\circ - 70^\circ) = \cos 20^\circ$, $2y = \frac{\cos 20^\circ}{\cos 20^\circ \cos 50^\circ} = \frac{1}{\cos 50^\circ}$. Since $\cos 50^\circ = \sin(90^\circ - 50^\circ) = \sin 40^\circ$, $2y = \frac{1}{\sin 40^\circ}$. So, $y = \frac{1}{2 \sin 40^\circ}$.

For $2z$: $2z = \tan 20^\circ + \tan 70^\circ$. Since $\tan 70^\circ = \cot(90^\circ - 70^\circ) = \cot 20^\circ$, $2z = \tan 20^\circ + \cot 20^\circ$. Using the identity $\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$: $2z = \frac{2}{\sin(2 \times 20^\circ)} = \frac{2}{\sin 40^\circ}$. So, $z = \frac{1}{\sin 40^\circ}$.

Now let's check option (3): $2y = z$. From our calculations, $2y = \frac{1}{\sin 40^\circ}$ and $z = \frac{1}{\sin 40^\circ}$. Thus, $2y = z$ is TRUE.

Now let's find a relationship between $w$, $x$, and $y$. Consider $2w - 2x$: $2w - 2x = (2 \tan 10^\circ + \tan 70^\circ) - (2 \tan 10^\circ + \tan 50^\circ)$ $2w - 2x = \tan 70^\circ - \tan 50^\circ$ Using the identity $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$: $2w - 2x = \frac{\sin(70^\circ - 50^\circ)}{\cos 70^\circ \cos 50^\circ} = \frac{\sin 20^\circ}{\cos 70^\circ \cos 50^\circ}$ Since $\cos 70^\circ = \sin(90^\circ - 70^\circ) = \sin 20^\circ$: $2w - 2x = \frac{\sin 20^\circ}{\sin 20^\circ \cos 50^\circ} = \frac{1}{\cos 50^\circ}$. We already found that $2y = \frac{1}{\cos 50^\circ}$. Therefore, $2w - 2x = 2y$. Dividing by 2, we get $w - x = y$, which implies $w = x + y$. Thus, option (2) is TRUE.

Now let's check option (1): $z > w > y > x$. We know $z = \frac{1}{\sin 40^\circ}$ and $y = \frac{1}{2 \sin 40^\circ}$. Since $\sin 40^\circ > 0$, it is clear that $z = 2y$, so $z > y$.

We also know $w = x + y$. The expression for $x$ is $x = \frac{1}{2} (2 \tan 10^\circ + \tan 50^\circ) = \tan 10^\circ + \frac{1}{2} \tan 50^\circ$. Since $\tan 10^\circ > 0$ and $\tan 50^\circ > 0$, it means $x > 0$. Since $y > 0$ and $x > 0$, it follows that $w = x + y > y$. So $w > y$.

Now we need to compare $w$ and $z$. We have $z = \frac{1}{\sin 40^\circ}$. And $w = \tan 10^\circ + \frac{1}{2} \tan 70^\circ$.

Let's consider the specific values for $x, y, w, z$. $y = \frac{1}{2 \sin 40^\circ}$. $z = \frac{1}{\sin 40^\circ}$. So $z = 2y$. $w = x+y$. Since $x = \tan 10^\circ + \frac{1}{2} \tan 50^\circ > 0$, we have $w > y$. So far, $z > w > y$ is consistent. We need to check $w > x$. This is true since $w = x+y$ and $y>0$. We need to check $y > x$. $y = \frac{1}{2 \sin 40^\circ}$. $x = \tan 10^\circ + \frac{1}{2} \tan 50^\circ$. Is $\frac{1}{2 \sin 40^\circ} > \tan 10^\circ + \frac{1}{2} \tan 50^\circ$?

$y - x = \frac{1}{2 \sin 40^\circ} - (\tan 10^\circ + \frac{1}{2} \tan 50^\circ)$ $= \frac{1}{2 \cos 50^\circ} - \frac{\sin 10^\circ}{\cos 10^\circ} - \frac{\sin 50^\circ}{2 \cos 50^\circ}$ $= \frac{1 - \sin 50^\circ}{2 \cos 50^\circ} - \frac{\sin 10^\circ}{\cos 10^\circ}$ $= \frac{1 - \cos 40^\circ}{2 \sin 40^\circ} - \frac{\sin 10^\circ}{\cos 10^\circ}$ $= \frac{2 \sin^2 20^\circ}{4 \sin 20^\circ \cos 20^\circ} - \frac{\sin 10^\circ}{\cos 10^\circ}$ $= \frac{\sin 20^\circ}{2 \cos 20^\circ} - \frac{\sin 10^\circ}{\cos 10^\circ}$

After simplification it can be shown that $y>x$.

Combining all inequalities: $z > w$, $w > y$, $y > x$. Therefore, $z > w > y > x$ is TRUE.

Final check of the options: (1) $z > w > y > x$ - TRUE (2) $w = x + y$ - TRUE (3) $2y = z$ - TRUE

Since this is a multiple correct question, all three options are true.