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The value of cot 12^(@) cot 38^(@) cot 5...

The value of `cot 12^(@) cot 38^(@) cot 52^(@) cot 60^(@) cot 78^(@)` is:

A

1

B

0

C

`1/sqrt(2)`

D

`1/sqrt(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the value of \( \cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ \), we can use the properties of cotangent and the complementary angles. ### Step-by-step Solution: 1. **Identify Complementary Angles**: We know that \( \cot(90^\circ - \theta) = \tan(\theta) \). This property will help us simplify the cotangent terms: - \( \cot 12^\circ = \cot(90^\circ - 78^\circ) = \tan 78^\circ \) - \( \cot 38^\circ = \cot(90^\circ - 52^\circ) = \tan 52^\circ \) Therefore, we can rewrite the expression as: \[ \cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ = \tan 78^\circ \tan 52^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ \] 2. **Simplify the Expression**: Now, substituting the values we have: \[ = \tan 78^\circ \cdot 1 \cdot \cot 60^\circ \cdot \cot 78^\circ \] We know that \( \cot 60^\circ = \frac{1}{\sqrt{3}} \). 3. **Using the Identity**: We also know that \( \tan 78^\circ = \frac{1}{\cot 78^\circ} \). Thus, we can rewrite: \[ = \tan 78^\circ \cdot \cot 78^\circ \cdot \cot 60^\circ \] Since \( \tan 78^\circ \cdot \cot 78^\circ = 1 \), we have: \[ = 1 \cdot \cot 60^\circ = \cot 60^\circ \] 4. **Final Calculation**: Now substituting the value of \( \cot 60^\circ \): \[ = \frac{1}{\sqrt{3}} \] Thus, the value of \( \cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ \) is \( \frac{1}{\sqrt{3}} \). ### Final Answer: \[ \frac{1}{\sqrt{3}} \]

To solve the problem of finding the value of \( \cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ \), we can use the properties of cotangent and the complementary angles. ### Step-by-step Solution: 1. **Identify Complementary Angles**: We know that \( \cot(90^\circ - \theta) = \tan(\theta) \). This property will help us simplify the cotangent terms: - \( \cot 12^\circ = \cot(90^\circ - 78^\circ) = \tan 78^\circ \) - \( \cot 38^\circ = \cot(90^\circ - 52^\circ) = \tan 52^\circ \) ...
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