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Question

If Tan θ = 7/24, then what is the value of p in (tanθ - secθ)/sinθ = -p/28 ?

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

75

Understanding the Trigonometry Problem

The question asks us to find the value of 'p' in a given trigonometric equation. We are provided with the value of $\tan \theta$ and a relationship involving $\tan \theta$, $\sec \theta$, and $\sin \theta$. To solve this, we first need to find the values of $\sin \theta$ and $\sec \theta$ using the given value of $\tan \theta$.

Finding Trigonometric Ratios from Tan $\theta$

Given that $\tan \theta = 7/24$. In a right-angled triangle, $\tan \theta$ is defined as the ratio of the opposite side to the adjacent side. So, we can consider the opposite side as 7 units and the adjacent side as 24 units.

Using the Pythagorean theorem, we can find the hypotenuse (h):

$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

$h^2 = 7^2 + 24^2$

$h^2 = 49 + 576$

$h^2 = 625$

$h = \sqrt{625}$

$h = 25$

Now that we have all three sides of the right triangle (opposite = 7, adjacent = 24, hypotenuse = 25), we can find $\sin \theta$ and $\cos \theta$:

  • $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$
  • $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$

Next, we can find $\sec \theta$, which is the reciprocal of $\cos \theta$:

  • $\sec \theta = \frac{1}{\cos \theta} = \frac{1}{24/25} = \frac{25}{24}$

Let's summarize the trigonometric ratios we found:

Ratio Value
$\tan \theta$ $7/24$
$\sin \theta$ $7/25$
$\cos \theta$ $24/25$
$\sec \theta$ $25/24$

Substituting Values into the Equation

The given equation is $(\tan \theta - \sec \theta) / \sin \theta = -p/28$. We will substitute the values we found for $\tan \theta$, $\sec \theta$, and $\sin \theta$ into this equation.

$\frac{\frac{7}{24} - \frac{25}{24}}{\frac{7}{25}} = \frac{-p}{28}$

Simplifying the Left Side of the Equation

First, simplify the numerator on the left side:

$\frac{7}{24} - \frac{25}{24} = \frac{7 - 25}{24} = \frac{-18}{24}$

The fraction $\frac{-18}{24}$ can be simplified by dividing both the numerator and denominator by their greatest common divisor, which is 6:

$\frac{-18 \div 6}{24 \div 6} = \frac{-3}{4}$

Now, substitute this back into the left side of the equation:

$\frac{-3/4}{7/25}$

To divide by a fraction, we multiply by its reciprocal:

$\frac{-3}{4} \times \frac{25}{7} = \frac{-3 \times 25}{4 \times 7} = \frac{-75}{28}$

Solving for p

Now the equation becomes:

$\frac{-75}{28} = \frac{-p}{28}$

To find the value of p, we can multiply both sides of the equation by 28:

$\frac{-75}{28} \times 28 = \frac{-p}{28} \times 28$

$-75 = -p$

Multiply both sides by -1:

$-75 \times (-1) = -p \times (-1)$

$75 = p$

So, the value of p is 75.

Verification with Options

The calculated value of p is 75, which matches one of the provided options.

Revision Table: Key Concepts

Concept Description Formula/Example
Tan $\theta$ Ratio of opposite side to adjacent side in a right triangle. $\tan \theta = \text{Opposite} / \text{Adjacent}$
Sin $\theta$ Ratio of opposite side to hypotenuse in a right triangle. $\sin \theta = \text{Opposite} / \text{Hypotenuse}$
Sec $\theta$ Reciprocal of Cos $\theta$. $\sec \theta = 1 / \cos \theta = \text{Hypotenuse} / \text{Adjacent}$
Pythagorean Theorem Relates the sides of a right triangle. $a^2 + b^2 = c^2$

Additional Information: Trigonometric Identities

Beyond the basic ratios derived from a right triangle, trigonometry involves various identities that relate these functions. Some fundamental identities are:

  • Reciprocal Identities: $\csc \theta = 1/\sin \theta$, $\sec \theta = 1/\cos \theta$, $\cot \theta = 1/\tan \theta$
  • Quotient Identities: $\tan \theta = \sin \theta / \cos \theta$, $\cot \theta = \cos \theta / \sin \theta$
  • Pythagorean Identities: $\sin^2 \theta + \cos^2 \theta = 1$, $1 + \tan^2 \theta = \sec^2 \theta$, $1 + \cot^2 \theta = \csc^2 \theta$

These identities are useful for simplifying expressions and solving more complex trigonometric equations.

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