CBSE Class 10 Chapter 8 Introduction of Trigonometry Exercise 8.2

 

Q1.

Evaluate the following : (i) $\sin 60^{0} \cos 30^{0} + \sin 30^{0} \cos 60^{0}$ (ii) $2\tan^{2} 45^{0} + \cos^{2} 30^{0}-\sin^{2} 60^{0}$ (iii) $\frac {\cos 45^{0}}{\sec 30^{0}+\text {cosec} 30^{0}}$ (iv) $\frac {\sin 30^{0}+\tan 45^{0}-\text {cosec}60^{0}}{\sec 30^{0}+\cos 60^{0}+cot 45^{0}}$ (v) $\frac {5\cos^{2}60^{0}+4\sec^{2}30^{0}-\tan^{2}45^{0}}{\sec^{2}30^{0}+\cos^{2}30^{0}}$

(i)

$\sin 60^{0} \cos 30^{0} + \sin 30^{0} \cos 60^{0}$

Ans.

We know that $\sin 60^\circ = \frac {\sqrt {3}}{2}$ $\cos 30^\circ = \frac {\sqrt {3}}{2}$ $\sin 30^\circ = \frac {1}{2}$ $\cos 60^\circ = \frac {1}{2}$ By putting all value, we get $$\begin{align*} \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ &= \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right) \\ &= \frac{3}{4} + \frac{1}{4} \\ &= 1 \end{align*}$$ Hence,  $\sin 60^{0} \cos 30^{0} + \sin 30^{0} \cos 60^{0}=1$

(ii)

$2\tan^{2} 45^{0} + \cos^{2} 30^{0}-\sin^{2} 60^{0}$

Ans.

We know that $\tan 45^\circ = 1$ $\cos 30^\circ = \frac {\sqrt {3}}{2}$ $\sin 60^\circ = \frac {\sqrt {3}}{2}$ By putting all value, we get $$\begin{align*} 2\tan^{2} 45^\circ + \cos^{2} 30^\circ - \sin^{2} 60^\circ & = 2 \cdot 1^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \\ &= 2 + \frac{3}{4} - \frac{3}{4} \\ &= 2. \end{align*}$$ Hence,  $2\tan^{2} 45^{0} + \cos^{2} 30^{0}-\sin^{2} 60^{0}=2$

(iii)

$\frac {\cos 45^{0}}{\sec 30^{0}+\text {cosec} 30^{0}}$

Ans.

We know that $\cos 45^\circ = \frac {{1}}{\sqrt {2}}$ $\sec 30^\circ = \frac {1}{\cos 30^{0}}= \frac {1}{\frac {\sqrt {3}}{2}}=\frac {2}{\sqrt {3}}$ $\text {cosec} 30^\circ = \frac {1}{\sin 30^{0}}= \frac{1}{\frac{1}{2}}=\frac {2}{1}=2$ By putting all value, we get $$\begin{align*} \frac {\cos 45^{0}}{\sec 30^{0}+\text {cosec} 30^{0}}&=\frac{\frac{1}{\sqrt{2}}}{\left(\frac{2}{\sqrt{3}}\right)+\left(\frac{2}{1}\right)}\\ &=\frac{1}{\sqrt{2}}\times \frac {1}{\left(\frac{2}{\sqrt{3}}\right)+\left(\frac{2}{1}\right)}\\ &=\frac{1}{\sqrt{2}}\times\frac{1}{\frac{2\times 1+2\times\sqrt{3}}{\sqrt{3}\times 1}}\\ &=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}\times 1}{2\times 1+2\times\sqrt{3}}\\ &=\frac{1}{\sqrt{2}}\times\frac{\sqrt {3}}{2+2\sqrt{3}}\\ &=\frac{1\times\sqrt{3}}{\sqrt{2}\times2(1+\sqrt{3})}\\ &=\frac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}\\ &=\frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}\\ &=\text {By rationalizing}\\ &=\frac{\sqrt{3}(\sqrt {3}-1)}{2\sqrt{2}(\sqrt{3}+1)(\sqrt {3}-1)}\\ &= \text {We know that} (a+b)(a-b)=a^{2}-b^{2}\text { So,}\\ &=\frac{3-\sqrt{3}}{2\sqrt{2}\left((\sqrt{3})^{2}-1^{2}\right)}\\ &=\frac{3-\sqrt{3}}{2\sqrt{2}\left(3-1\right)}\\ &=\frac{3-\sqrt{3}}{2\sqrt{2}\times 2}\\ &=\frac{(3-\sqrt{3})}{4\sqrt{2}}\\ &=\text {Multiply Numerator and Denominator by }\sqrt {2}\\ &=\frac{(3-\sqrt{3})}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ &=\frac{3\sqrt{2}-\sqrt{3}\times \sqrt{2}}{4\times 2}\\ &=\frac{3\sqrt{2}-\sqrt{6}}{8}\\ \end{align*}$$ Hence,  $\frac {\cos 45^{0}}{\sec 30^{0}+\text {cosec} 30^{0}}=\frac{3\sqrt{2}-\sqrt{6}}{8}$

(iv)

$\frac {\sin 30^{0}+\tan 45^{0}-\text {cosec}60^{0}}{\sec 30^{0}+\cos 60^{0}+cot 45^{0}}$

Ans.

We know that $\sin 30^\circ = \frac {1}{2}$ $\tan 45^\circ = 1$ $\text {cosec } 60^\circ = \frac {2}{\sqrt {3}}$ $\sec 30^\circ =\frac {2}{\sqrt {3}}$ $\cos 60^\circ = \frac {1}{2}$ $\cot 45^\circ = 1$ By putting all value, we get $$\begin{align*} &=\frac {\frac{1}{2}+1-\frac{2}{\sqrt {3}}}{\frac {2}{\sqrt{3}}+\frac {1}{2}+1}\\ &= \frac {\frac {\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac {4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\\ &=\frac {\sqrt{3}+2\sqrt{3}-4}{4+\sqrt{3}+2\sqrt{3}}\\ &=\frac {3\sqrt{3}-4}{4+3\sqrt{3}}\\ &=\text {by rationalizing}\\ &=\frac {3\sqrt{3}-4}{4+3\sqrt{3}}\times \frac {4-3\sqrt{3}}{4-3\sqrt{3}}\\ &=\frac{(3\sqrt{3}-4)\times (4-3\sqrt{3})}{(4+3\sqrt{3})\times (4-3\sqrt{3})}\\ &=\frac {12\sqrt{3}-27-16+12\sqrt{3}}{16-12\sqrt{3}+12\sqrt{3}-27}\\ &=\frac {12\sqrt{3}-27-16+12\sqrt{3}}{-11}\\ &=\frac {24\sqrt{3}-43}{-11}\\ &=\frac {43-24\sqrt{3}}{-11} \end{align*}$$ Hence,  $\frac {\cos 45^{0}}{\sec 30^{0}+\text {cosec} 30^{0}}=\frac {43-24\sqrt{3}}{-11}$

(v)

$\frac {5\cos^{2}60^{0}+4\sec^{2}30^{0}-\tan^{2}45^{0}}{\sec^{2}30^{0}+\cos^{2}30^{0}}$

Ans.

We know that $\cos 60^\circ = \frac {1}{2}$ $\sec 30^\circ =\frac {2}{\sqrt {3}}$ $\tan 45^\circ = 1$ $\cos 30^\circ = \frac {\sqrt {3}}{2}$ By putting all value, we get $$\begin{align*} \frac {5\cos^{2}60^{0}+4\sec^{2}30^{0}-\tan^{2}45^{0}}{\sec^{2}30^{0}+\cos^{2}30^{0}} &=\frac {5\times \left(\frac {1}{2}\right)^{2}+4\times \left(\frac {2}{\sqrt{3}}\right)^{2}-1}{\left(\frac{1}{2}\right)^{2}+\left(\frac {\sqrt{3}}{2}\right)^{2}}\\ &=\frac {5\times\frac {1}{4}+4\times\frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}\\ &=\frac {\frac {5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}\\ &=\frac {\frac {15+64-12}{12}}{\frac{1+3}{4}}\\ &=\frac {15+64-12}{12}\times 1\\ &=\bf \frac {67}{12} \end{align*}$$ Hence, $\frac {5\cos^{2}60^{0}+4\sec^{2}30^{0}-\tan^{2}45^{0}}{\sec^{2}30^{0}+\cos^{2}30^{0}}=\bf \frac {67}{12}$

Q2.	Choose the correct option and justify your choice :

(i)

$\frac{2\tan30^{0}}{1+\tan^{2}30^{0}}=$ $\text {(A) } \sin60^{0} \text {       (B)} \cos60^{0} \text {       (C)} \tan 60^{0} \text {       (D)} \sin 30^{0}$

Ans.

$$\begin{align*} \frac{2\tan30^{0}}{1+\tan^{2}30^{0}} &=\frac {2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}\\ &=\frac {\frac{2}{\sqrt{3}}}{\frac{1}{1}+\frac{1}{3}}\\ &=\frac {\frac{2}{\sqrt{3}}}{\frac{3+1}{3}}\\ &=\frac {2}{\sqrt{3}}\times \frac {3}{4}\\ &=\frac {3}{2\sqrt{3}}\times \frac {\sqrt{3}}{\sqrt{3}}\\ &=\frac {3\sqrt{3}}{2\times 3}\\ &=\frac {\sqrt{3}}{2}=\sin 60^{0} \end{align*}$$ Therefore, Correct Option is (A)

(ii)	$\frac{1-\tan^{2}45^{0}}{1+\tan^{2}45^{0}}=$ $\text {(A) } \tan90^{0} \text {       (B)} 1 \text {       (C)} \sin 45^{0} \text {       (D)} 0$
Ans.	$$\begin{align*} \frac{1-\tan^{2}45^{0}}{1+\tan^{2}45^{0}} &=\frac {1-(1)^{2}}{1+(1)^{2}}\\ &=\frac {1-1}{1+1}\\ &=\frac{0}{2}\\ &=0 \end{align*}$$ Therefore, Correct Option is (D)

(iii)	$\sin 2A=2\sin A \text { is true when A=}$ $\text {(A)} 0^{0} \text {(B)} 30^{0} \text {(C)} 45^{0} \text {(D)} 60^{0}$
Ans.	$\sin 2A=2\sin A \text {, for} A=0^{0}\text { Then}$ $\text {L.H.S.}=\sin 2A = \sin 2\times 0^{0}=\sin 0^{0} = 0^{0}$ $\text {R.H.S.}=2\sin A= 2\sin 0^{0}=2\times 0=0$ Therefore, Correct Option is (A)

(iv)

$\frac {2\tan 30^{0}}{1-\tan^{2}30^{0}}=$ $\text {(A)} \cos 60^{0} \text {(B)} \sin 60^{0} \text {(C)} \tan 60^{0} \text {(D)} \sin 30^{0}$

Ans.

$$\begin{align*} \frac {2\tan 30^{0}}{1-\tan^{2}30^{0}} &=\frac {2\left(\frac {1}{\sqrt{3}}\right)}{1-\left(\frac {1}{\sqrt{3}}\right)^{2}}\\ &=\frac {\frac {2}{\sqrt{3}}}{\frac{1}{1}-\frac {1}{3}}\\ &=\frac {\frac {2}{\sqrt{3}}}{\frac{3-1}{3}}\\ &=\frac {2}{\sqrt{3}}\times \frac {3}{2}\\ &=\frac {3}{\sqrt{3}}\times \frac {\sqrt{3}}{\sqrt {3}}\\ &=\frac {3\sqrt{3}}{3}\\ &=\sqrt {3}=\tan 60^{0} \end{align*}$$ Therefore, Correct Option is (C)

Q3.

If $\tan (A+B)=\sqrt {3}$ and $\tan (A-B)=\frac {1}{\sqrt {3}}$ ; $0^{0}<A+B \leq 90^{0}$ ; $A>B$, Find A and B.

Ans.

$\tan \left(A+B\right)=\sqrt {3}=\tan 60^{0}$ $\Rightarrow A+B = 60^{0}...........(i)$ $\tan \left(A-B\right)=\frac {1}{\sqrt {3}}=\tan 30^{0}$ $\Rightarrow A-B = 30^{0}...........(ii)$ $\text {Adding equation (i) and (ii)}$ $A+B+A-B=60^{0}+30^{0}$ $2A=90^{0}$ $A=\frac {90}{2}=45^{0}$ $\text {Putting value in equation (i)}$ $45^{0}+B = 60^{0}\Rightarrow B=15^{0}$ Hence, $A = 45^{0}$ and $B=15^{0}$

Q4.

State whether the following are true or false. Justify your answer. (i) $\sin (A + B) = sin A + sin B$. (ii) The value of $\sin\theta$ increases as $\theta$ increases. (iii) The value of $\cos\theta$ increases as $\theta$ increases. (iv) $\sin\theta = \cos\theta$ for all values of $\theta$. (v) $\cot A$ is not defined for $A = 0^{0}$.

Ans.

(i) $\sin (A + B) = sin A + sin B$. Let's take $A=60^{0}$ and $B=30^{0}$, then By putting value $\text {LHS} = \sin (A+B) = \sin (60^{0}+30^{0})=\sin 90^{0}=1$ $\text {RHS} = \sin A + \sin B = \sin 60^{0}+\sin 30^{0}=\frac {3}{2}+\frac{1}{2}=1+\frac {\sqrt{3}}{2}$ Because value are not equal,so given statement are False. (ii) The value of $\sin\theta$ increases as $\theta$ increases. We know that the values obtained from unit circle for sin are: $\sin 0^{0}=0$ $\sin 30^{0}=\frac {1}{2}$ $\sin 45^{0}=\frac {1}{\sqrt {2}}$ $\sin 60^{0}=\frac {\sqrt{3}}{2}$ $\sin 90^{0}=1$ Therefore, If the value of $\sin\theta$ increases as $\theta$ increases, so the given statement is True. (iii) The value of $\cos\theta$ increases as $\theta$ increases. We know that the values obtained from unit circle for $\cos$ are: $\cos 0^{0}=1$ $\cos 30^{0}=\frac {\sqrt{3}}{2}$ $\cos 45^{0}=\frac {1}{\sqrt {2}}$ $\cos 60^{0}=\frac {1}{2}$ $\cos 90^{0}=0$ Therefore, If the value of $\cos\theta$ decreases as $\theta$ increases, so the given statement is False. (iv) $\sin\theta = \cos\theta$ for all values of $\theta$ The sine and cosine functions in degrees are defined based on the unit circle.  $\sin\theta =\cos\theta$ is possible, only when a right angle triangle has 2 angle of $\frac {\pi}{4}$. so, given statement is False. (v) $\cot A$ is not defined for $A = 0^{0}$ We know that- $\cot A =\frac {\cos A}{\sin A}$ For $A=0^{0}$, then $\sin A = \sin 0^{0}=0$ amd division by zero is undefined. Therefore $\cot 0^{0}$ is not defined for $A=0^{0}$. so, given statement is True.