NCERT Solutions Class 10 Maths Chapter 8: Introduction to Trigonometry

#### **Q1. Evaluate the following:** **(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$** **(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$** **(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ}$** **(iv) $\frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$** **(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$** #### **A1. Solution:** **(i)** $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}) + (\frac{1}{2} \times \frac{1}{2}) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = \mathbf{1}$. **(ii)** $2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2 = 2 + \frac{3}{4} - \frac{3}{4} = \mathbf{2}$. **(iii)** $\frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{1/\sqrt{2}}{(2 + 2\sqrt{3})/\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$ After rationalizing the denominator: $\mathbf{\frac{3\sqrt{2} - \sqrt{6}}{8}}$. **(iv)** $\frac{1/2 + 1 - 2/\sqrt{3}}{2/\sqrt{3} + 1/2 + 1} = \frac{3/2 - 2/\sqrt{3}}{3/2 + 2/\sqrt{3}} = \frac{(3\sqrt{3}-4)/2\sqrt{3}}{(3\sqrt{3}+4)/2\sqrt{3}} = \frac{3\sqrt{3}-4}{3\sqrt{3}+4}$ After rationalizing: $\mathbf{\frac{43 - 24\sqrt{3}}{11}}$. **(v)** $\frac{5(1/2)^2 + 4(2/\sqrt{3})^2 - (1)^2}{(1/2)^2 + (\sqrt{3}/2)^2} = \frac{5/4 + 16/3 - 1}{1/4 + 3/4} = \frac{(15 + 64 - 12)/12}{1} = \mathbf{\frac{67}{12}}$. #### **Q2. Choose the correct option and justify your choice:** **(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$** (A) $\sin 60^\circ$ (B) $\cos 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$ **(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$** (A) $\tan 90^\circ$ (B) $1$ (C) $\sin 45^\circ$ (D) $0$ **(iii) $\sin 2A = 2 \sin A$ is true when A =** (A) $0^\circ$ (B) $30^\circ$ (C) $45^\circ$ (D) $60^\circ$ **(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} =$** (A) $\cos 60^\circ$ (B) $\sin 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$ #### **A2. Solution:** **(i)** $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{\sqrt{3}}{2} = \sin 60^\circ$. **Ans: (A)** **(ii)** $\frac{1 - 1^2}{1 + 1^2} = \frac{0}{2} = 0$. **Ans: (D)** **(iii)** If $A = 0^\circ$: LHS = $\sin(2 \times 0) = \sin 0 = 0$; RHS = $2 \sin 0 = 2 \times 0 = 0$. **Ans: (A)** **(iv)** $\frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} = \tan 60^\circ$. **Ans: (C)** #### **Q3. If $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ; A > B$, find A and B.** #### **A3. Solution:** $\tan(A + B) = \sqrt{3} \Rightarrow \tan(A + B) = \tan 60^\circ \Rightarrow A + B = 60^\circ$ (1) $\tan(A - B) = \frac{1}{\sqrt{3}} \Rightarrow \tan(A - B) = \tan 30^\circ \Rightarrow A - B = 30^\circ$ (2) Adding (1) and (2): $2A = 90^\circ \Rightarrow A = 45^\circ$. Substitute $A$ in (1): $45^\circ + B = 60^\circ \Rightarrow B = 15^\circ$. **Ans: $A = 45^\circ, B = 15^\circ$.** #### **Q4. State whether the following are true or false. Justify your answer.** **(i) $\sin (A + B) = \sin A + \sin B$** **False.** Let $A=30^\circ, B=60^\circ$. $\sin(90^\circ)=1$, but $\sin 30^\circ + \sin 60^\circ = 1/2 + \sqrt{3}/2 \neq 1$. **(ii) The value of $\sin \theta$ increases as $\theta$ increases.** **True.** In the interval $[0^\circ, 90^\circ]$, value goes from $0$ to $1$. **(iii) The value of $\cos \theta$ increases as $\theta$ increases.** **False.** Value decreases from $1$ to $0$ as $\theta$ increases from $0^\circ$ to $90^\circ$. **(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.** **False.** Only true at $\theta = 45^\circ$. For $\theta = 30^\circ, \sin 30^\circ = 1/2$ and $\cos 30^\circ = \sqrt{3}/2$. **(v) $\cot A$ is not defined for $A = 0^\circ$.** **True.** $\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$, which is undefined.