Solving the Trigonometry Problem: Finding sin θ + cos θ - tan θ

The problem asks us to find the value of the expression $\sin \theta + \cos \theta - \tan \theta$ given that $\text{cosec } \theta = \frac{13}{12}$.

We know that the cosecant of an angle is the reciprocal of the sine of the angle. So, if $\text{cosec } \theta = \frac{13}{12}$, then we can easily find $\sin \theta$:

$\sin \theta = \frac{1}{\text{cosec } \theta} = \frac{1}{\frac{13}{12}} = \frac{12}{13}$.

Now we have the value of $\sin \theta$. To find $\cos \theta$ and $\tan \theta$, we can use trigonometric identities or visualize a right-angled triangle.

Let's use the concept of a right-angled triangle. For an acute angle $\theta$ in a right triangle:

• $\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}}$
• $\cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}}$
• $\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}}$
• $\text{cosec } \theta = \frac{\text{Hypotenuse}}{\text{Opposite Side}}$

Given $\text{cosec } \theta = \frac{13}{12}$, we can consider a right triangle where the Hypotenuse is 13 units and the Opposite Side (to angle $\theta$) is 12 units.

We need to find the length of the Adjacent Side. We can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let the Adjacent Side be 'a'. According to the Pythagorean theorem:

$\text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2$

$12^2 + a^2 = 13^2$

$144 + a^2 = 169$

$a^2 = 169 - 144$

$a^2 = 25$

$a = \sqrt{25} = 5$

So, the Adjacent Side is 5 units.

Now we have the lengths of all three sides of the right triangle:

• Opposite Side = 12
• Adjacent Side = 5
• Hypotenuse = 13

We can now find the values of $\cos \theta$ and $\tan \theta$:

$\cos \theta = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{5}{13}$

$\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{12}{5}$

Now we have all the required trigonometric ratios:

• $\sin \theta = \frac{12}{13}$
• $\cos \theta = \frac{5}{13}$
• $\tan \theta = \frac{12}{5}$

Substitute these values into the expression $\sin \theta + \cos \theta - \tan \theta$:

$\sin \theta + \cos \theta - \tan \theta = \frac{12}{13} + \frac{5}{13} - \frac{12}{5}$

First, add the fractions with the same denominator:

$\frac{12}{13} + \frac{5}{13} = \frac{12 + 5}{13} = \frac{17}{13}$

Now, subtract $\frac{12}{5}$ from $\frac{17}{13}$:

$\frac{17}{13} - \frac{12}{5}$

To subtract these fractions, find a common denominator. The least common multiple (LCM) of 13 and 5 is $13 \times 5 = 65$.

Convert each fraction to have a denominator of 65:

$\frac{17}{13} = \frac{17 \times 5}{13 \times 5} = \frac{85}{65}$

$\frac{12}{5} = \frac{12 \times 13}{5 \times 13} = \frac{156}{65}$

Now perform the subtraction:

$\frac{85}{65} - \frac{156}{65} = \frac{85 - 156}{65} = \frac{-71}{65}$

So, the value of $\sin \theta + \cos \theta - \tan \theta$ is $-\frac{71}{65}$.

Revision Table: Key Trigonometric Concepts

Additional Information: Trigonometric Identities

Besides the reciprocal identities, there are other fundamental trigonometric identities that are very useful in solving problems. Some important ones include:

• Pythagorean Identities:
  • $\sin^2 \theta + \cos^2 \theta = 1$
  • $1 + \tan^2 \theta = \sec^2 \theta$
  • $1 + \cot^2 \theta = \text{cosec}^2 \theta$
• Quotient Identities:
  • $\tan \theta = \frac{\sin \theta}{\cos \theta}$
  • $\cot \theta = \frac{\cos \theta}{\sin \theta}$

These identities can be used as alternatives to the right-triangle method to find missing trigonometric ratios if one ratio is known.

For example, knowing $\sin \theta = \frac{12}{13}$, you could use $\sin^2 \theta + \cos^2 \theta = 1$ to find $\cos \theta$:

$(\frac{12}{13})^2 + \cos^2 \theta = 1$

$\frac{144}{169} + \cos^2 \theta = 1$

$\cos^2 \theta = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$

$\cos \theta = \sqrt{\frac{25}{169}} = \frac{5}{13}$ (assuming $\theta$ is in a quadrant where cosine is positive).

Then, $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{12/13}{5/13} = \frac{12}{5}$. This matches the results from the triangle method.