In ΔPQR, right-angled at Q, PQ = 12 cm and PR = 13 cm. Find the value of tan P - cot R.

First, QR = √(13² - 12²) = 5 cm. tan P = QR/PQ = 5/12 and cot R = QR/PQ = 5/12. Therefore, tan P - cot R = 5/12 - 5/12 = 0.

If sin A = 3/4, calculate cos A and tan A.

Let perpendicular = 3k and hypotenuse = 4k. Base = √(16k² - 9k²) = √7k. Thus, cos A = Base/Hypotenuse = √7/4 and tan A = Perpendicular/Base = 3/√7.

If cot θ = 7/8, evaluate (i) (1 + sin θ)(1 - sin θ) / (1 + cos θ)(1 - cos θ) (ii) cot² θ.

(i) The expression simplifies to (1 - sin² θ) / (1 - cos² θ) = cos² θ / sin² θ = cot² θ. (ii) Since cot θ = 7/8, cot² θ = (7/8)² = 49/64. Both parts result in 49/64.

State whether the statement 'The value of tan A is always less than 1' is true or false.

False. The value of tan A can be greater than 1, as the opposite side can be longer than the adjacent side (e.g., tan 60° = √3 ≈ 1.732).

NCERT Solutions Class 10 Maths Chapter 8: Introduction to Trigonometry

Q: In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C.

By Pythagoras theorem, AC = 25 cm. (i) sin A = BC/AC = 7/25, cos A = AB/AC = 24/25. (ii) sin C = AB/AC = 24/25, cos C = BC/AC = 7/25.

Q: In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

By Pythagoras theorem and given PR + QR = 25, we find QR = 12 cm and PR = 13 cm. Thus, sin P = 12/13, cos P = 5/13, and tan P = 12/5.

Detailed solutions for Exercise 8.1 (Trigonometric Ratios).

Ex 8.1 Ex 8.2 Ex 8.3

#### **Q1. In $\Delta ABC$, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:** **(i) $\sin A, \cos A$** **(ii) $\sin C, \cos C$** #### **A1. Solution:** By Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$. $AC = \sqrt{625} = 25$ cm. **(i) For $\angle A$:** Opposite side = $BC = 7$, Adjacent side = $AB = 24$, Hypotenuse = $25$. $\sin A = \frac{BC}{AC} = \frac{7}{25}$ $\cos A = \frac{AB}{AC} = \frac{24}{25}$ **(ii) For $\angle C$:** Opposite side = $AB = 24$, Adjacent side = $BC = 7$, Hypotenuse = $25$. $\sin C = \frac{AB}{AC} = \frac{24}{25}$ $\cos C = \frac{BC}{AC} = \frac{7}{25}$ #### **Q2. In Fig. 8.13, find $\tan P - \cot R$.** *(Given: $PQ = 12$ cm, $PR = 13$ cm, $\angle Q = 90^\circ$)* #### **A2. Solution:** In $\Delta PQR$, $QR^2 = PR^2 - PQ^2 = 13^2 - 12^2 = 169 - 144 = 25$. $QR = 5$ cm. $\tan P = \frac{\text{Opposite to } P}{\text{Adjacent to } P} = \frac{QR}{PQ} = \frac{5}{12}$ $\cot R = \frac{\text{Adjacent to } R}{\text{Opposite to } R} = \frac{QR}{PQ} = \frac{5}{12}$ $\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0$. #### **Q3. If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.** #### **A3. Solution:** Let $\sin A = \frac{3k}{4k}$ where $k$ is a positive constant. Opposite side = $3k$, Hypotenuse = $4k$. Adjacent side = $\sqrt{(4k)^2 - (3k)^2} = \sqrt{16k^2 - 9k^2} = \sqrt{7}k$. $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}$ $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}$ #### **Q4. Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.** #### **A4. Solution:** $\cot A = \frac{8}{15} = \frac{\text{Adjacent}}{\text{Opposite}}$. Let Adjacent = $8k$, Opposite = $15k$. Hypotenuse = $\sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$. $\sin A = \frac{15k}{17k} = \frac{15}{17}$ $\sec A = \frac{17k}{8k} = \frac{17}{8}$ #### **Q5. Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.** #### **A5. Solution:** $\sec \theta = \frac{13}{12} = \frac{\text{Hypotenuse}}{\text{Adjacent}}$. Let Hypotenuse = $13k$, Adjacent = $12k$. Opposite = $\sqrt{(13k)^2 - (12k)^2} = \sqrt{169k^2 - 144k^2} = \sqrt{25k^2} = 5k$. $\sin \theta = \frac{5}{13}, \quad \cos \theta = \frac{12}{13}, \quad \tan \theta = \frac{5}{12}$ $\text{cosec } \theta = \frac{13}{5}, \quad \cot \theta = \frac{12}{5}$ #### **Q6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.** #### **A6. Solution:** Consider a right triangle $ABC$ (right-angled at $C$ or two triangles $CAD$ and $CBE$). In $\Delta ABC$ (where $C$ is $90^\circ$): $\cos A = \frac{AC}{AB}$, $\cos B = \frac{BC}{AB}$. Given $\cos A = \cos B \Rightarrow \frac{AC}{AB} = \frac{BC}{AB}$. $\Rightarrow AC = BC$. In a triangle, angles opposite to equal sides are equal. Therefore, $\angle B = \angle A$. #### **Q7. If $\cot \theta = \frac{7}{8}$, evaluate:** **(i) $\frac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}$** **(ii) $\cot^2 \theta$** #### **A7. Solution:** **(i)** Expression = $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta$. Given $\cot \theta = \frac{7}{8}$, so $\cot^2 \theta = (\frac{7}{8})^2 = \frac{49}{64}$. **(ii)** $\cot^2 \theta = (\frac{7}{8})^2 = \frac{49}{64}$. #### **Q8. If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.** #### **A8. Solution:** $3 \cot A = 4 \Rightarrow \cot A = \frac{4}{3} \Rightarrow \tan A = \frac{3}{4}$. Adjacent = $4k$, Opposite = $3k \Rightarrow$ Hypotenuse = $5k$. $\sin A = \frac{3}{5}, \cos A = \frac{4}{5}$. LHS: $\frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$. RHS: $(\frac{4}{5})^2 - (\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$. **LHS = RHS. Yes, they are equal.** #### **Q9. In $\Delta ABC$, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:** **(i) $\sin A \cos C + \cos A \sin C$** **(ii) $\cos A \cos C - \sin A \sin C$** #### **A9. Solution:** $\tan A = \frac{1}{\sqrt{3}} \Rightarrow \text{Opposite} (BC) = 1k, \text{Adjacent} (AB) = \sqrt{3}k$. $AC = \sqrt{(\sqrt{3}k)^2 + (1k)^2} = \sqrt{3k^2 + k^2} = 2k$. $\sin A = \frac{1}{2}, \cos A = \frac{\sqrt{3}}{2}, \sin C = \frac{\sqrt{3}}{2}, \cos C = \frac{1}{2}$. **(i)** $(\frac{1}{2})(\frac{1}{2}) + (\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) = \frac{1}{4} + \frac{3}{4} = 1$. **(ii)** $(\frac{\sqrt{3}}{2})(\frac{1}{2}) - (\frac{1}{2})(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$. #### **Q10. In $\Delta PQR$, right-angled at Q, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.** #### **A10. Solution:** Let $QR = x$. Then $PR = 25 - x$. $PR^2 = PQ^2 + QR^2$ $(25 - x)^2 = 5^2 + x^2$ $625 - 50x + x^2 = 25 + x^2$ $600 = 50x \Rightarrow x = 12$. So, $QR = 12$ cm, $PR = 13$ cm, $PQ = 5$ cm. $\sin P = \frac{12}{13}, \cos P = \frac{5}{13}, \tan P = \frac{12}{5}$. #### **Q11. State whether the following are true or false. Justify your answer.** **(i) The value of $\tan A$ is always less than 1.** **False.** $\tan A$ can be any value (e.g., $\tan 60^\circ = \sqrt{3} \approx 1.732$). **(ii) $\sec A = \frac{12}{5}$ for some value of angle A.** **True.** Secant must be $\ge 1$. $12/5 = 2.4$, which is possible. **(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.** **False.** $\cos A$ is for cosine; $\text{cosec } A$ is for cosecant. **(iv) $\cot A$ is the product of $\cot$ and $A$.** **False.** It is a single trigonometric function of the angle $A$. **(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.** **False.** Sine value cannot be greater than 1 ($4/3 = 1.33$).