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Question

The value of $\frac{\sqrt{3}\operatorname{cosec}20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}$ is equal to

The correct answer is
16

Trigonometric Expression Solution

We need to evaluate the given trigonometric expression:

$ E = \frac{\sqrt{3}\operatorname{cosec}20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ} $

Numerator Calculation

Let the numerator be $N = \sqrt{3}\operatorname{cosec}20^\circ - \sec 20^\circ$.

Expressing in terms of $\sin$ and $\cos$:

$ N = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} $

Find a common denominator:

$ N = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} $

Multiply the numerator and denominator by 2 to use trigonometric identities:

$ N = \frac{2(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ)}{2\sin 20^\circ \cos 20^\circ} $

Apply the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$ to the numerator, with $A=60^\circ$ and $B=20^\circ$ ($\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 60^\circ = \frac{1}{2}$):

$ N = \frac{2\sin(60^\circ - 20^\circ)}{\sin(2 \times 20^\circ)} $

Simplify using the double angle formula for sine ($\sin(2A) = 2\sin A \cos A$):

$ N = \frac{2\sin 40^\circ}{\sin 40^\circ} $

$ N = 4 $

Denominator Calculation

Let the denominator be $D = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$.

We know the value $\cos 60^\circ = \frac{1}{2}$.

$ D = \frac{1}{2} (\cos 20^\circ \cos 40^\circ \cos 80^\circ) $

Use the product identity $\cos x \cos(60^\circ - x) \cos(60^\circ + x) = \frac{1}{4}\cos(3x)$.

Here, $x = 20^\circ$, so $60^\circ - x = 40^\circ$ and $60^\circ + x = 80^\circ$.

$ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4}\cos(3 \times 20^\circ) = \frac{1}{4}\cos 60^\circ $

Substitute the value of $\cos 60^\circ = \frac{1}{2}$:

$ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} $

Now, substitute this back into the expression for $D$:

$ D = \frac{1}{2} \times \frac{1}{8} = \frac{1}{16} $

Final Value Calculation

The value of the expression $E$ is the numerator divided by the denominator:

$ E = \frac{N}{D} = \frac{4}{\frac{1}{16}} $

$ E = 4 \times 16 = 64 $

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Similar Questions

  1. Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.

  2. The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
  3. If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha$, then $\tan \alpha$ is equal to _________.
  4. Let $\vec{a_k} = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k - 1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a_k}|^2}{\sum_{k=1}^n |\vec{b_k}|^2}$ is _____.

  5. If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
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    Statement p : 

    The value of $sin120^\circ$ can be derived by taking $\theta = 240^\circ$ in the equation $2sin \frac{\theta}{2} = \sqrt{1+sin \theta} - \sqrt{1-sin \theta}$ 

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Important Questions from Trigonometry

  1. Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.

  2. The number of elements in the set $\{x \in [0, 180^\circ] : \tan(x + 100^\circ) = \tan(x + 50^\circ) \tan x \tan(x - 50^\circ)\}$ is _______.
  3. If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha$, then $\tan \alpha$ is equal to _________.
  4. Let $\vec{a_k} = (\tan \theta_k) \hat{i} + \hat{j}$ and $\vec{b_k} = \hat{i} - (\cot \theta_k) \hat{j}$, where $\theta_k = \frac{2^{k - 1}\pi}{2^n + 1}$, for some $n \in \mathbb{N}, n > 5$. Then the value of $\frac{\sum_{k=1}^n |\vec{a_k}|^2}{\sum_{k=1}^n |\vec{b_k}|^2}$ is _____.

  5. If $k = \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)\right) + \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)\right)$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is ________
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