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Introduction To Trigonometry
EXERCISE 8.1
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
In Fig. 8.13, find tan P – cot R.
Step-by-Step Solution
Key Concept: In a right‑angled triangle the two acute angles are complementary (their sum is 90°). For complementary angles θ and (90° − θ), the identity tan θ = cot (90° − θ) holds. Hence tan P and cot R are equal because ∠R = 90° − ∠P.
1. From Fig. 8.13 we observe that \(\triangle PQR\) is a right‑angled triangle with the right angle at \(Q\). Therefore the two acute angles satisfy
$$\angle P + \angle R = 90^{\circ}.$$
2. Hence \(\angle R = 90^{\circ} - \angle P\).
3. Using the complementary‑angle identity
$$\cot R = \cot\bigl(90^{\circ} - \angle P\bigr) = \tan \angle P,$$
because \(\cot(90^{\circ} - \theta) = \tan \theta\).
4. Therefore
$$\tan P - \cot R = \tan P - \tan P = 0.$$
5. The required value is \(0\).
$$\angle P + \angle R = 90^{\circ}.$$
2. Hence \(\angle R = 90^{\circ} - \angle P\).
3. Using the complementary‑angle identity
$$\cot R = \cot\bigl(90^{\circ} - \angle P\bigr) = \tan \angle P,$$
because \(\cot(90^{\circ} - \theta) = \tan \theta\).
4. Therefore
$$\tan P - \cot R = \tan P - \tan P = 0.$$
5. The required value is \(0\).
Correct Answer: 0
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