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.