Evaluating the Trigonometric Expression

The problem asks us to evaluate the following expression:

$$ 7\left(\frac{\text{cosec } 24^\circ}{\text{sec } 66^\circ}\right)^3 + 8\left(\frac{\text{cot } 37^\circ}{\text{tan } 53^\circ}\right)^4 - \left(2\frac{\text{sec } 14^\circ}{\text{cosec } 76^\circ}\right)^2 + \left(-3\frac{\text{tan } 82^\circ}{\text{cot } 8^\circ}\right)^3 $$

To solve this, we will use the trigonometric identities related to complementary angles. Recall that for complementary angles ($ \alpha $ and $ \beta $) where $ \alpha + \beta = 90^\circ $, the following relationships hold:

• $ \text{cosec } \alpha = \text{sec } \beta $
• $ \text{sec } \alpha = \text{cosec } \beta $
• $ \text{cot } \alpha = \text{tan } \beta $
• $ \text{tan } \alpha = \text{cot } \beta $

Step-by-Step Evaluation

Term 1 Evaluation: $ 7\left(\frac{\text{cosec } 24^\circ}{\text{sec } 66^\circ}\right)^3 $

We observe that $ 24^\circ + 66^\circ = 90^\circ $. Therefore, the angles are complementary.

Using the identity $ \text{sec } \beta = \text{cosec } (90^\circ - \beta) $, we have $ \text{sec } 66^\circ = \text{cosec } (90^\circ - 66^\circ) = \text{cosec } 24^\circ $.

Substituting this into the fraction:

$$ \frac{\text{cosec } 24^\circ}{\text{sec } 66^\circ} = \frac{\text{cosec } 24^\circ}{\text{cosec } 24^\circ} = 1 $$

So, the first term becomes:

$$ 7(1)^3 = 7 \times 1 = 7 $$

Term 2 Evaluation: $ 8\left(\frac{\text{cot } 37^\circ}{\text{tan } 53^\circ}\right)^4 $

Here, $ 37^\circ + 53^\circ = 90^\circ $. The angles are complementary.

Using the identity $ \text{tan } \beta = \text{cot } (90^\circ - \beta) $, we have $ \text{tan } 53^\circ = \text{cot } (90^\circ - 53^\circ) = \text{cot } 37^\circ $.

Substituting this into the fraction:

$$ \frac{\text{cot } 37^\circ}{\text{tan } 53^\circ} = \frac{\text{cot } 37^\circ}{\text{cot } 37^\circ} = 1 $$

So, the second term becomes:

$$ 8(1)^4 = 8 \times 1 = 8 $$

Term 3 Evaluation: $ -\left(2\frac{\text{sec } 14^\circ}{\text{cosec } 76^\circ}\right)^2 $

We note that $ 14^\circ + 76^\circ = 90^\circ $. The angles are complementary.

Using the identity $ \text{cosec } \beta = \text{sec } (90^\circ - \beta) $, we have $ \text{cosec } 76^\circ = \text{sec } (90^\circ - 76^\circ) = \text{sec } 14^\circ $.

Substituting this into the fraction:

$$ \frac{\text{sec } 14^\circ}{\text{cosec } 76^\circ} = \frac{\text{sec } 14^\circ}{\text{sec } 14^\circ} = 1 $$

So, the third term becomes:

$$ -\left(2 \times 1\right)^2 = -(2)^2 = -4 $$

Term 4 Evaluation: $ \left(-3\frac{\text{tan } 82^\circ}{\text{cot } 8^\circ}\right)^3 $

Here, $ 82^\circ + 8^\circ = 90^\circ $. The angles are complementary.

Using the identity $ \text{cot } \beta = \text{tan } (90^\circ - \beta) $, we have $ \text{cot } 8^\circ = \text{tan } (90^\circ - 8^\circ) = \text{tan } 82^\circ $.

Substituting this into the fraction:

$$ \frac{\text{tan } 82^\circ}{\text{cot } 8^\circ} = \frac{\text{tan } 82^\circ}{\text{tan } 82^\circ} = 1 $$

So, the fourth term becomes:

$$ \left(-3 \times 1\right)^3 = (-3)^3 = -27 $$

Combining the Results

Now, we add the evaluated terms together:

$$ \text{Total Value} = 7 + 8 - 4 + (-27) $$

$$ \text{Total Value} = 15 - 4 - 27 $$

$$ \text{Total Value} = 11 - 27 $$

$$ \text{Total Value} = -16 $$

Thus, the value of the given trigonometric expression is -16.