Finding the Value of cot 15° cot 20° cot 70° cot 75°

The problem asks us to find the value of the trigonometric expression: $\cot 15^\circ \cot 20^\circ \cot 70^\circ \cot 75^\circ$.

To solve this, we can use the trigonometric identity relating cotangent and tangent for complementary angles. The identity is:

$\cot (90^\circ - \theta) = \tan \theta$

Also, we know that $\tan \theta = \frac{1}{\cot \theta}$, which means $\cot \theta \cdot \tan \theta = 1$.

Let's look at the angles in the given expression:

• $15^\circ$ and $75^\circ$ are complementary angles because $15^\circ + 75^\circ = 90^\circ$.
• $20^\circ$ and $70^\circ$ are complementary angles because $20^\circ + 70^\circ = 90^\circ$.

Now, let's apply the identity $\cot (90^\circ - \theta) = \tan \theta$ to $\cot 70^\circ$ and $\cot 75^\circ$:

• For $\cot 70^\circ$: Since $70^\circ = 90^\circ - 20^\circ$, we have $\cot 70^\circ = \cot (90^\circ - 20^\circ) = \tan 20^\circ$.
• For $\cot 75^\circ$: Since $75^\circ = 90^\circ - 15^\circ$, we have $\cot 75^\circ = \cot (90^\circ - 15^\circ) = \tan 15^\circ$.

Now substitute these values back into the original expression:

The expression $\cot 15^\circ \cot 20^\circ \cot 70^\circ \cot 75^\circ$ becomes:

$\cot 15^\circ \cot 20^\circ (\tan 20^\circ) (\tan 15^\circ)$

We can rearrange the terms to group the cotangent and tangent of the same angle together:

$(\cot 15^\circ \tan 15^\circ) (\cot 20^\circ \tan 20^\circ)$

Using the identity $\cot \theta \cdot \tan \theta = 1$, we can simplify the expression:

• $\cot 15^\circ \tan 15^\circ = 1$
• $\cot 20^\circ \tan 20^\circ = 1$

So, the expression simplifies to:

$(1) \cdot (1) = 1$

Therefore, the value of $\cot 15^\circ \cot 20^\circ \cot 70^\circ \cot 75^\circ$ is 1.

The steps are summarized below:

The final value is 1.