Understanding the Trigonometry Problem

The problem asks us to find the value of a given trigonometric expression: \(\rm \frac{sin\,\theta(1-tan\,\theta)}{tan\,\theta(1+cosec\,\theta)}\), given that \(\cos \theta = \frac{12}{13}\). To solve this, we first need to find the values of \(\sin \theta\), \(\tan \theta\), and \(\operatorname{cosec} \theta\) using the given value of \(\cos \theta\).

Finding Trigonometric Ratios from cos θ

We are given \(\cos \theta = \frac{12}{13}\). We know that in a right-angled triangle, \(\cos \theta\) is the ratio of the adjacent side to the hypotenuse. Let's assume a right triangle where the adjacent side is 12 units and the hypotenuse is 13 units.

Using the Pythagorean theorem, which states that \(\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2\), we can find the length of the opposite side:

Let the opposite side be \(p\), the adjacent side be \(b\), and the hypotenuse be \(h\). We have \(b = 12\) and \(h = 13\).

\(p^2 + b^2 = h^2\)

\(p^2 + 12^2 = 13^2\)

\(p^2 + 144 = 169\)

\(p^2 = 169 - 144\)

\(p^2 = 25\)

\(p = \sqrt{25}\)

\(p = 5\)

So, the opposite side is 5 units.

Now we can find the other required trigonometric ratios:

• sin θ: Ratio of opposite side to hypotenuse. \(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}\)
• tan θ: Ratio of opposite side to adjacent side. \(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12}\)
• cosec θ: Reciprocal of \(\sin \theta\). \(\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\)

Evaluating the Trigonometric Expression

Now we substitute the values of \(\sin \theta\), \(\tan \theta\), and \(\operatorname{cosec} \theta\) into the given expression:

\(\rm \frac{sin\,\theta(1-tan\,\theta)}{tan\,\theta(1+cosec\,\theta)}\)

First, let's evaluate the numerator:

\(\sin \theta (1 - \tan \theta) = \frac{5}{13} \left(1 - \frac{5}{12}\right)\)

To subtract inside the parenthesis, find a common denominator (12):

\(\frac{5}{13} \left(\frac{12}{12} - \frac{5}{12}\right) = \frac{5}{13} \left(\frac{12-5}{12}\right) = \frac{5}{13} \left(\frac{7}{12}\right)\)

Multiply the fractions:

\(\frac{5 \times 7}{13 \times 12} = \frac{35}{156}\)

So, the numerator is \(\frac{35}{156}\).

Next, let's evaluate the denominator:

\(\tan \theta (1 + \operatorname{cosec} \theta) = \frac{5}{12} \left(1 + \frac{13}{5}\right)\)

To add inside the parenthesis, find a common denominator (5):

\(\frac{5}{12} \left(\frac{5}{5} + \frac{13}{5}\right) = \frac{5}{12} \left(\frac{5+13}{5}\right) = \frac{5}{12} \left(\frac{18}{5}\right)\)

Multiply the fractions:

\(\frac{5 \times 18}{12 \times 5}\)

We can cancel out the 5 from the numerator and denominator:

\(\frac{18}{12}\)

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 6:

\(\frac{18 \div 6}{12 \div 6} = \frac{3}{2}\)

So, the denominator is \(\frac{3}{2}\).

Finally, divide the numerator by the denominator:

Value of expression = \(\frac{\frac{35}{156}}{\frac{3}{2}}\)

Dividing by a fraction is the same as multiplying by its reciprocal:

\(\frac{35}{156} \times \frac{2}{3}\)

Multiply the numerators and denominators:

\(\frac{35 \times 2}{156 \times 3}\)

We can simplify by dividing 2 and 156 by 2:

\(\frac{35 \times 1}{78 \times 3}\)

Now, multiply the remaining terms:

\(\frac{35}{234}\)

Thus, the value of the expression \(\rm \frac{sin\,\theta(1-tan\,\theta)}{tan\,\theta(1+cosec\,\theta)}\) is \(\frac{35}{234}\).

Summary of Steps

• Use the given \(\cos \theta\) value and Pythagorean theorem to find the sides of a right triangle.
• Calculate \(\sin \theta\), \(\tan \theta\), and \(\operatorname{cosec} \theta\) using the side lengths.
• Substitute these values into the given trigonometric expression.
• Simplify the numerator and the denominator separately.
• Divide the simplified numerator by the simplified denominator to get the final value.

Revision Table: Trigonometric Ratios

Additional Information: Trigonometric Identities

Besides using a right-angled triangle, we can also use trigonometric identities to find the required ratios. For example, we know:

• \(\sin^2 \theta + \cos^2 \theta = 1\). From \(\cos \theta = \frac{12}{13}\), we get \(\sin^2 \theta + \left(\frac{12}{13}\right)^2 = 1\). This gives \(\sin^2 \theta + \frac{144}{169} = 1\). So, \(\sin^2 \theta = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}\). Thus, \(\sin \theta = \sqrt{\frac{25}{169}} = \frac{5}{13}\) (assuming θ is in a quadrant where sin is positive).
• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Using the values we found, \(\tan \theta = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}\).
• \(\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\).

These identity-based calculations confirm the ratios found using the right-angled triangle method and can be very useful in solving trigonometric problems.