What is the value of (tan 10° tan 80° + tan 20° tan 70° + tan 30° tan 60° + tan 40° tan 50°)?
4
Solving the Tangent Product Sum Problem
The question asks us to find the value of the expression: $ (\tan 10^\circ \tan 80^\circ + \tan 20^\circ \tan 70^\circ + \tan 30^\circ \tan 60^\circ + \tan 40^\circ \tan 50^\circ) $. This problem involves trigonometric ratios of complementary angles.
Understanding Complementary Angles in Trigonometry
Two angles are called complementary if their sum is $90^\circ$. For complementary angles, certain trigonometric identities hold true. One important identity we will use here is:
- $ \tan(90^\circ - \theta) = \cot(\theta) $
Also, we know the reciprocal relationship between tangent and cotangent:
- $ \tan(\theta) \cdot \cot(\theta) = 1 $ (provided $ \tan(\theta) $ and $ \cot(\theta) $ are defined)
Applying Identities to Each Term
Let's look at each term in the given expression and apply the complementary angle identity:
- Term 1: $ \tan 10^\circ \tan 80^\circ $
Here, $10^\circ + 80^\circ = 90^\circ$, so $10^\circ$ and $80^\circ$ are complementary angles.
We can write $ \tan 80^\circ $ as $ \tan (90^\circ - 10^\circ) $.
Using the identity, $ \tan (90^\circ - 10^\circ) = \cot 10^\circ $.
So, $ \tan 10^\circ \tan 80^\circ = \tan 10^\circ \cot 10^\circ $.
Using the reciprocal identity, $ \tan 10^\circ \cot 10^\circ = 1 $. - Term 2: $ \tan 20^\circ \tan 70^\circ $
Here, $20^\circ + 70^\circ = 90^\circ$.
We can write $ \tan 70^\circ $ as $ \tan (90^\circ - 20^\circ) $.
Using the identity, $ \tan (90^\circ - 20^\circ) = \cot 20^\circ $.
So, $ \tan 20^\circ \tan 70^\circ = \tan 20^\circ \cot 20^\circ $.
Using the reciprocal identity, $ \tan 20^\circ \cot 20^\circ = 1 $. - Term 3: $ \tan 30^\circ \tan 60^\circ $
Here, $30^\circ + 60^\circ = 90^\circ$.
We can write $ \tan 60^\circ $ as $ \tan (90^\circ - 30^\circ) $.
Using the identity, $ \tan (90^\circ - 30^\circ) = \cot 30^\circ $.
So, $ \tan 30^\circ \tan 60^\circ = \tan 30^\circ \cot 30^\circ $.
Using the reciprocal identity, $ \tan 30^\circ \cot 30^\circ = 1 $. - Term 4: $ \tan 40^\circ \tan 50^\circ $
Here, $40^\circ + 50^\circ = 90^\circ$.
We can write $ \tan 50^\circ $ as $ \tan (90^\circ - 40^\circ) $.
Using the identity, $ \tan (90^\circ - 40^\circ) = \cot 40^\circ $.
So, $ \tan 40^\circ \tan 50^\circ = \tan 40^\circ \cot 40^\circ $.
Using the reciprocal identity, $ \tan 40^\circ \cot 40^\circ = 1 $.
Calculating the Total Sum
Now we substitute the value of each term back into the original expression:
$ (\tan 10^\circ \tan 80^\circ) + (\tan 20^\circ \tan 70^\circ) + (\tan 30^\circ \tan 60^\circ) + (\tan 40^\circ \tan 50^\circ) $
$ = 1 + 1 + 1 + 1 $
$ = 4 $
The value of the given expression is 4.
| Term | Complementary Angle Identity Used | Result |
|---|---|---|
| $ \tan 10^\circ \tan 80^\circ $ | $ \tan 80^\circ = \cot 10^\circ $ | $ \tan 10^\circ \cot 10^\circ = 1 $ |
| $ \tan 20^\circ \tan 70^\circ $ | $ \tan 70^\circ = \cot 20^\circ $ | $ \tan 20^\circ \cot 20^\circ = 1 $ |
| $ \tan 30^\circ \tan 60^\circ $ | $ \tan 60^\circ = \cot 30^\circ $ | $ \tan 30^\circ \cot 30^\circ = 1 $ |
| $ \tan 40^\circ \tan 50^\circ $ | $ \tan 50^\circ = \cot 40^\circ $ | $ \tan 40^\circ \cot 40^\circ = 1 $ |
Final Answer for Tangent Product Sum
Summing up the results for each term, the total value is $1 + 1 + 1 + 1 = 4$. Therefore, the value of $ (\tan 10^\circ \tan 80^\circ + \tan 20^\circ \tan 70^\circ + \tan 30^\circ \tan 60^\circ + \tan 40^\circ \tan 50^\circ) $ is 4.
Revision Table: Trigonometric Identities
| Identity | Description |
|---|---|
| $ \tan(90^\circ - \theta) = \cot(\theta) $ | Tangent of an angle is the cotangent of its complementary angle. |
| $ \cot(90^\circ - \theta) = \tan(\theta) $ | Cotangent of an angle is the tangent of its complementary angle. |
| $ \sin(90^\circ - \theta) = \cos(\theta) $ | Sine of an angle is the cosine of its complementary angle. |
| $ \cos(90^\circ - \theta) = \sin(\theta) $ | Cosine of an angle is the sine of its complementary angle. |
| $ \sec(90^\circ - \theta) = \csc(\theta) $ | Secant of an angle is the cosecant of its complementary angle. |
| $ \csc(90^\circ - \theta) = \sec(\theta) $ | Cosecant of an angle is the secant of its complementary angle. |
| $ \tan(\theta) \cdot \cot(\theta) = 1 $ | Tangent and cotangent are reciprocals. |
| $ \sin(\theta) \cdot \csc(\theta) = 1 $ | Sine and cosecant are reciprocals. |
| $ \cos(\theta) \cdot \sec(\theta) = 1 $ | Cosine and secant are reciprocals. |
Additional Information on Tangent and Cotangent of Complementary Angles
The relationship $ \tan(90^\circ - \theta) = \cot(\theta) $ is fundamental when dealing with trigonometric ratios of complementary angles. This relationship arises directly from the definitions of tangent and cotangent in a right-angled triangle. Consider a right-angled triangle ABC, right-angled at B. Let $ \angle BAC = \theta $. Then $ \angle BCA = 90^\circ - \theta $.
- $ \tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{AB} $
- $ \cot(\theta) = \frac{\text{Adjacent side}}{\text{Opposite side}} = \frac{AB}{BC} $
Now consider the angle $90^\circ - \theta$:
- $ \tan(90^\circ - \theta) = \frac{\text{Opposite side to }(90^\circ - \theta)}{\text{Adjacent side to }(90^\circ - \theta)} = \frac{AB}{BC} $
- $ \cot(90^\circ - \theta) = \frac{\text{Adjacent side to }(90^\circ - \theta)}{\text{Opposite side to }(90^\circ - \theta)} = \frac{BC}{AB} $
Comparing the ratios, we see that $ \tan(90^\circ - \theta) = \frac{AB}{BC} = \cot(\theta) $. This confirms the identity used to solve the problem. Similarly, $ \cot(90^\circ - \theta) = \frac{BC}{AB} = \tan(\theta) $. The problem leverages these identities by pairing tangent values of complementary angles, which simplifies the product to 1.
Important Questions from Trigonometric Functions
If \(\tan \alpha=\frac{1}{7}\), \(\sin \beta=\frac{1}{\sqrt{10}}\); \(0<\alpha, \beta<\frac{\pi}{2}\), then what is the value of cos (α + 2β) ?
What is the period of the function?
What is the value of p + q?
What is the value of pq?
What is pq equal to ?
