Question

tan10 tan40 tan50 tan80​

Answer 1

Answer:

Sin 18°/cos 72° + √3 [tan10 tan 30 tan 40 tan 50 tan 80]. =Sin 18°/cos(90-72) + √ 3 [tan10 tan 30 tan 40 tan(90-40) tan(90-10)].

Answer 2

Topic

Trigonometry (Acute angles)

Solution

Given Expression

$=\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80 ^{\circ}$

$=\tan (90^{\circ} -80^{\circ} )\tan(90^{\circ} -50^{\circ} )\tan 50^{\circ} \tan 80 ^{\circ}$

$=\dfrac{1}{\tan 80^{\circ} } \times \dfrac{1}{\tan 50^{\circ} } \times \tan 50^{\circ} \tan 80^{\circ}$

$=\boxed{1}$

Learn More

Why does $\tan 10^{\circ} =\dfrac{1}{\tan 80^{\circ} }$?

① In a right triangle…

Here is the first reason.

Suppose we have a right triangle with an angle of $10^{\circ}$. Then the remaining angle is $80^{\circ}$.

If $\tan 80^{\circ}=\dfrac{\text{Opposite}}{\text{Adjacent}}$, $\tan 10^{\circ} =\dfrac{\text{Adjacent}}{\text{Opposite}}$, so $\tan 10^{\circ}=\dfrac{1}{\tan 80^{\circ}}$.

② To the extent… (Class 11)

However, this explanation works in the range of acute angles. To the extent, we can use a unit circle and show symmetry.

The mathematical convention of the unit circle extends trigonometry to the range of other angles. If an angle formed by the positive x-axis, and a ray are $\theta$, we can define three values as $\sin \theta = \dfrac{y}{r} ,\ \cos \theta = \dfrac{x}{r} ,\ \tan \theta=\dfrac{y}{x}$. $[1]$

And this is the convention of trigonometric functions.

③ The result.

As the two angles are complementary, angles are symmetric against $y=x$. $[2]$

$\tan \theta = \dfrac{x}{y}$ and $\tan (\dfrac{\pi }{2} -\theta )=\dfrac{y}{x}$.

Hence, $\tan (\dfrac{\pi }{2} -\theta )=\dfrac{1}{\tan \theta}$.

Information

$[1]$ Attachment included.

$[2]$ Attachment included.