Question

$((\sqrt{2}-1)cosec \theta - sec \theta)cos3\theta$ at $\theta = \frac{\pi}{24}$ is-

• A. 0
• B. -2
• C. 1
• D. $\sqrt{3}$

Answer 1

The correct answer is 2, which is not present in the options. There might be an error in the question or options.

Answer 2

To evaluate the expression $((\sqrt{2}-1)cosec \theta - sec \theta)cos3\theta$ at $\theta = \frac{\pi}{24}$, we proceed as follows:

1. Substitute $\theta = \frac{\pi}{24}$ into the expression.
2. Simplify $cos(3\theta) = cos(\frac{3\pi}{24}) = cos(\frac{\pi}{8})$.
3. Rewrite $cosec \theta$ as $\frac{1}{\sin \theta}$ and $sec \theta$ as $\frac{1}{\cos \theta}$.
4. Combine the terms inside the parenthesis: $\frac{(\sqrt{2}-1)\cos \theta - \sin \theta}{\sin \theta \cos \theta}$.
5. Recognize that $\sqrt{2}-1 = \tan(\frac{\pi}{8})$.
6. Substitute $\tan(\frac{\pi}{8})$ into the numerator and simplify using trigonometric identities, specifically the sine subtraction formula.
7. Simplify the denominator using the double angle identity for sine.
8. Cancel out common terms to arrive at the final result.

The step-by-step solution is as follows:

Given expression: $((\sqrt{2}-1)cosec \theta - sec \theta)cos3\theta$

Substitute $\theta = \frac{\pi}{24}$:

$((\sqrt{2}-1)cosec \frac{\pi}{24} - sec \frac{\pi}{24})cos(3 \cdot \frac{\pi}{24}) = ((\sqrt{2}-1)cosec \frac{\pi}{24} - sec \frac{\pi}{24})cos(\frac{\pi}{8})$

Rewrite in terms of sine and cosine:

$\left( \frac{\sqrt{2}-1}{\sin \frac{\pi}{24}} - \frac{1}{\cos \frac{\pi}{24}} \right) \cos(\frac{\pi}{8})$

Combine terms:

$\frac{(\sqrt{2}-1)\cos \frac{\pi}{24} - \sin \frac{\pi}{24}}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Use $\sqrt{2}-1 = \tan(\frac{\pi}{8})$:

$\frac{\tan(\frac{\pi}{8})\cos \frac{\pi}{24} - \sin \frac{\pi}{24}}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Rewrite $\tan(\frac{\pi}{8})$ as $\frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}$:

$\frac{\frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}\cos \frac{\pi}{24} - \sin \frac{\pi}{24}}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Simplify the numerator:

$\frac{\sin(\frac{\pi}{8})\cos \frac{\pi}{24} - \cos(\frac{\pi}{8})\sin \frac{\pi}{24}}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$\frac{\sin(\frac{\pi}{8} - \frac{\pi}{24})}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Simplify the angle: $\frac{\pi}{8} - \frac{\pi}{24} = \frac{3\pi - \pi}{24} = \frac{2\pi}{24} = \frac{\pi}{12}$:

$\frac{\sin(\frac{\pi}{12})}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\sin \frac{\pi}{24} \cos \frac{\pi}{24}} \cos(\frac{\pi}{8})$

Using the double angle identity $\sin(2x) = 2\sin x \cos x$, we have $\sin x \cos x = \frac{1}{2}\sin(2x)$:

$\frac{\sin(\frac{\pi}{12})}{\cos(\frac{\pi}{8})} \cdot \frac{1}{\frac{1}{2}\sin(\frac{\pi}{12})} \cos(\frac{\pi}{8})$

Simplify:

$\frac{\sin(\frac{\pi}{12})}{\cos(\frac{\pi}{8})} \cdot \frac{2}{\sin(\frac{\pi}{12})} \cdot \cos(\frac{\pi}{8})$

Cancel common terms:

$2$

Therefore, the value of the expression is 2, which is not among the given options. There may be an error in the question or the options provided.