Given sec = 13 , 12 calculate all other trigonometric ratios.
Step-by-Step Solution
Key Concept: Use the reciprocal identities (\(\sec\theta = \dfrac{1}{\cos\theta}\), \(\csc\theta = \dfrac{1}{\sin\theta}\), \(\cot\theta = \dfrac{1}{\tan\theta}\)) and the fundamental Pythagorean identity \(\sin^{2}\theta + \cos^{2}\theta = 1\) to find the remaining ratios.
1. Find \(\cos\theta\) using the reciprocal identity of secant:
$$\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\dfrac{13}{12}} = \frac{12}{13}.$$
2. Find \(\sin\theta\) from the Pythagorean identity:
$$\sin^{2}\theta = 1 - \cos^{2}\theta = 1 - \left(\frac{12}{13}\right)^{2} = 1 - \frac{144}{169} = \frac{25}{169}.$$
Hence,
$$\sin\theta = \sqrt{\frac{25}{169}} = \frac{5}{13}$$ (taking the positive value as \(\theta\) is an acute angle in the context of NCERT exercises).
3. Find \(\tan\theta\) using the definition \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\):
$$\tan\theta = \frac{\dfrac{5}{13}}{\dfrac{12}{13}} = \frac{5}{12}.$$
4. Find \(\csc\theta\) (reciprocal of sine):
$$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{\dfrac{5}{13}} = \frac{13}{5}.$$
5. Find \(\cot\theta\) (reciprocal of tangent):
$$\cot\theta = \frac{1}{\tan\theta} = \frac{1}{\dfrac{5}{12}} = \frac{12}{5}.$$
6. Verify the given secant (reciprocal of cosine):
$$\sec\theta = \frac{1}{\cos\theta} = \frac{1}{\dfrac{12}{13}} = \frac{13}{12},$$
which matches the given value, confirming the calculations.
Correct Answer: {\(\sin\theta = \dfrac{5}{13}\), \(\cos\theta = \dfrac{12}{13}\), \(\tan\theta = \dfrac{5}{12}\), \(\csc\theta = \dfrac{13}{5}\), \(\sec\theta = \dfrac{13}{12}\), \(\cot\theta = \dfrac{12}{5}\)}