EXERCISE 8.2
1. Evaluate the following :
(i) \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
Solution: We use the known trigonometric values:
\( \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}, \)
\( \cos 60^\circ = \frac{1}{2} \)
Substituting these into the expression:
\( = \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{2} \cdot \frac{1}{2} \right) \)
\( = \frac{3}{4} + \frac{1}{4} \)
\( = \frac{4}{4} \)
\( = 1 \)
(ii) \(2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ\)
Solution: We use the known trigonometric values
\(\tan 45^\circ = 1 \Rightarrow \tan^2 45^\circ = 1^2 = 1\)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \Rightarrow \cos^2 30^\circ = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \)
\(\sin 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 60^\circ = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4}\)
Now substitute the values into the expression then we get
\(2 \cdot 1 + \frac{3}{4} – \frac{3}{4} = 2 + 0 = 2\)
(iii) \(\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}\)
Solution: We have,
\( \cos 45^\circ = \frac{1}{\sqrt{2}}\)
\(\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\)
\(\csc 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{\frac{1}{2}} = 2\)
Now substituting this values in the given equation
\(\left( \frac{1}{\sqrt{2}} \right) \cdot \left( \frac{2}{\sqrt{3}} \right) + 2\)
\(= \frac{2}{\sqrt{2} \cdot \sqrt{3}} + 2\)
\(= \frac{2}{\sqrt{6}} + 2\)
\(= \frac{2\sqrt{6}}{6} + 2\)
\(= \frac{\sqrt{6}}{3} + 2\)
(iv) \(\frac{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}\)
Solution: Given,
\(\frac{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}\)
We have,
\(\sin 30^\circ = \frac{1}{2}\)
\(\tan 45^\circ = 1\)
\(\csc 60^\circ = \frac{2}{\sqrt{3}}\)
\(\sec 30^\circ = \frac{2}{\sqrt{3}}\)
\(\cos 60^\circ = \frac{1}{2}\)
\(\cot 45^\circ = 1\)
Now,
Substituting these values in given expression then we get,
\(=\frac{\frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}\)
\(\text{Numerator=} \frac{3}{2} – \frac{2}{\sqrt{3}} \)
\(\frac{3}{2} – \frac{2}{\sqrt{3}} = \frac{9 – 4\sqrt{3}}{6}\)
\(\text{Denominator=} \frac{3}{2} + \frac{2}{\sqrt{3}}\)
\(\frac{3}{2} + \frac{2}{\sqrt{3}} = \frac{9 + 4\sqrt{3}}{6}\)
now,
\(\frac{Numerator}{Denominator}= \frac{\frac{9 – 4\sqrt{3}}{6}}{\frac{9 + 4\sqrt{3}}{6}}\)
\(\Rightarrow\frac{\frac{9 – 4\sqrt{3}}{6}}{\frac{9 + 4\sqrt{3}}{6}} = \frac{9 – 4\sqrt{3}}{9 + 4\sqrt{3}}\)
\(\Rightarrow\frac{9 – 4\sqrt{3}}{9 + 4\sqrt{3}} \cdot \frac{9 – 4\sqrt{3}}{9 – 4\sqrt{3}}\)
\(\text{Denominator=}9 + 4\sqrt{3})(9 – 4\sqrt{3}) \)
\( = 9^2 – (4\sqrt{3})^2 = 81 – 48 = 33\)
\(\text{Numerator=}9 – 4\sqrt{3})^2 = 9^2 – 2 \cdot 9 \cdot 4\sqrt{3} + (4\sqrt{3})^2 \)
\(= 81 – 72\sqrt{3} + 48 = 129 – 72\sqrt{3}\)
\(\Rightarrow\frac{(9 – 4\sqrt{3})^2}{(9 + 4\sqrt{3})(9 – 4\sqrt{3})} \)
\( = \frac{129 – 72\sqrt{3}}{33}\)
(v) \(\frac{5\cos 60^\circ + 4\sec 30^\circ – \tan 45^\circ}{\sin 30^\circ \cdot \cos 30^\circ}\)
Solution: Given, \(\frac{5\cos 60^\circ + 4\sec 30^\circ – \tan 45^\circ}{\sin 30^\circ \cdot \cos 30^\circ}\)
We have,
\(\cos 60^\circ = \frac{1}{2}\)
\(\sec 30^\circ = \frac{2}{\sqrt{3}}\)
\(\tan 45^\circ = 1\)
\(\sin 30^\circ = \frac{1}{2}\)
\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
Now,
\(\Rightarrow\frac{5 \cdot \frac{1}{2} + 4 \cdot \frac{2}{\sqrt{3}} – 1}{\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}=
\frac{\frac{5}{2} + \frac{8}{\sqrt{3}} – 1}{\frac{\sqrt{3}}{4}}\)
\(\Rightarrow\frac{5}{2} – 1 = \frac{3}{2}, \text{so numerator becomes } \frac{3}{2} + \frac{8}{\sqrt{3}}\)
\(\Rightarrow\frac{\frac{3}{2} + \frac{8}{\sqrt{3}}}{\frac{\sqrt{3}}{4}} = \left( \frac{3}{2} + \frac{8}{\sqrt{3}} \right) \cdot \frac{4}{\sqrt{3}}\)
\(\Rightarrow\frac{3}{2} \cdot \frac{4}{\sqrt{3}} + \frac{8}{\sqrt{3}} \cdot \frac{4}{\sqrt{3}} = \frac{12}{\sqrt{3}} + \frac{32}{3}\)
\(\Rightarrow\frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\)
\(\Rightarrow 4\sqrt{3} + \frac{32}{3}\)
2.Choose the correct option and justify your choice :
(i) \(\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}\)
Solution: We know that,
\(\tan 30^\circ = \frac{1}{\sqrt{3}}, \quad \tan^2 30^\circ = \frac{1}{3}\)
Substitute into the expression:
\( \frac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} \)
\( = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} \)
\( = \frac{3\sqrt{3}}{2 \cdot 3} = \frac{\sqrt{3}}{2} \)
Correct Option: \( \text{(A)}\quad \sin 60^\circ = \frac{\sqrt{3}}{2} \)
(ii) \(\frac{1 – \tan^245^\circ}{1 + \tan^245^\circ}\)
Solution: Given, \(\frac{1 – \tan^245^\circ}{1 + \tan^245^\circ}\)
we know that
\(\tan(45^\circ) = 1\)
Substituting this value in given expression we get
\(\frac{1 – \tan^2(45^\circ)}{1 + \tan^2(45^\circ)} = \frac{1 – 1^2}{1 + 1^2} = \frac{1 – 1}{1 + 1} = \frac{0}{2} = 0\)
Correct option is \( (D)=0\)
(iii) \(\sin(2A) = 2\sin(A)\) is true when A=?
Solution: We have, \(\sin(2A) = 2\sin(A)\)
Use the double angle identity:
\(\sin(2A) = 2\sin(A)\cos(A)\)
So:
\(2sin(A)\cos(A) = 2\sin(A)\)
Divide both sides by 2:
\(\sin(A)\cos(A) = \sin(A)\)
Subtract \(\sin(A)\) from both sides:
\(\sin(A)\cos(A) – \sin(A) = 0\)
Factor out \(sin(A)\):
\(\sin(A)(\cos(A) – 1) = 0\)
This gives two possible cases:
\(\sin(A) = 0 \Rightarrow A = 0^\circ, 180^\circ, \ldots)\)
\(\cos(A) = 1 \Rightarrow A = 0^\circ, 360^\circ, \ldots)\)
Correct option is: \(A = 0^\circ\)
(iv) \(\frac{2 \tan(30^\circ)}{1 – \tan^2(30^\circ)}=\)
Solution: We have, \(\frac{2 \tan(30^\circ)}{1 – \tan^2(30^\circ)}\)
We use the double angle identity for tangent:
\(\tan(2A) = \frac{2 \tan A}{1 – \tan^2 A}\)
Let \( A = 30^\circ \). Then:
\(\frac{2 \tan(30^\circ)}{1 – \tan^2(30^\circ)} = \tan(2 \cdot 30^\circ) = \tan(60^\circ)\)
Correct option is (C) \(=tan(60^\circ)\)
3. If \(\tan(A + B) = \sqrt{3}\) and \(\tan(A – B) = \frac{1}{\sqrt{3}}\): \(0^\circ < A + B \leq 90^\circ, A > B\), find A and B.
Solution: Given,\(\tan(A + B) = \sqrt{3}\) and \(\tan(A – B) = \frac{1}{\sqrt{3}}\)
we know that,
\(\tan(60^\circ) = \sqrt{3} \Rightarrow A + B = 60^\circ\)
\(\tan(30^\circ) = \frac{1}{\sqrt{3}} \Rightarrow A – B = 30^\circ\)
\(A + B = 60^\circ \)
\(A – B = 30^\circ\)
Add these equations:
\(2A = 90^\circ \Rightarrow A = 45^\circ\)
subtract these equations:
\(2B = 30^\circ \Rightarrow B = 15^\circ\)
therefore, \(A = 45^\circ, and \quad B = 15^\circ\)
4. State whether the following are true or false. Justify your answer:
(i) \(\sin (A + B) = sin A + sin B\)
Ans: False
Justification:
Let \( A = B = \frac{\pi}{4}\).
Left-hand side (LHS):
\(\sin(A + B) = \sin\left(\frac{\pi}{4} + \frac{\pi}{4}\right) \)
\( = \sin\left(\frac{\pi}{2}\right) = 1\)
Right-hand side (RHS):
\(\sin A + \sin B = \sin\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) \)
\( = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414\)
Therefore,
\(\sin(A + B) \ne \sin A + \sin B\)
(ii) The value of \( \sin \theta\) increases as \(\theta\) increases.
Ans: False.
Justification: The sine function does not increase continuously as θ increases. It increases and decreases periodically.
note: Behavior of \(\sin\theta\)
From \(\theta = 0^\circ\) to \( \theta = 90^\circ\ sin \theta \) does increase.
\(\sin 0^\circ = 0\)
\(\sin 30^\circ = 0.5\)
\(\sin 90^\circ = 1\)
From \(\theta = 90^\circ ) to ( \theta = 180^\circ\), the sine function starts decreasing:
\(\sin 120^\circ = \frac{\sqrt{3}}{2} \approx 0.866\)
\(\sin 150^\circ = 0.5\)
\(\sin 180^\circ = 0\)
Then, it becomes negative from \(80^\circ \) to \(270^\circ\),and increases again up to \(360^\circ\).
(iii) The value of \(\cos\theta\) increases as \(\theta\) increases.
Ans: False.
Justification: The value of cos θ does not always increase as θ increases. It decreases in some intervals and increases in others due to its wave-like, periodic nature.
note: Behavior of \(\cos\theta\)
From \( \theta = 0^\circ\quad\ to \quad \theta = 90^\circ\), \( \cos \theta\) decreases:
\(\cos 0^\circ = 1\)
\(\cos 60^\circ = 0.5\)
\(\cos 90^\circ = 0\)
From \( \theta = 90^\circ\quad\ to\quad \theta = 180^\circ \), \( \cos \theta \) continues to decrease, and becoming negative:
\(\cos 120^\circ = -0.5\)
\(\cos 180^\circ = -1\)
Then from \( \theta = 180^\circ\quad\ to \quad\theta = 360^\circ \), \( \cos \theta \) increases again:
\(cos 270^\circ = 0\)
\(cos 360^\circ = 1\)
(iv) \(\sin \theta = \cos \theta\) for all values of \(\theta\)
Ans: False
Justification: The sine and cosine functions are not equal for all values of \(\theta\).
(v) \(cot A\) is not defined for \(A = 0^\circ\)
Ans: True
Justification: The cotangent function is defined as:
\(\cot A = \frac{\cos A}{\sin A}\)
At \( A = 0^\circ \):
\(\cos 0^\circ = 1, \quad \sin 0^\circ = 0\)
\(\cot 0^\circ = \frac{1}{0}\)
Since division by zero is undefined, \(cot A\) is not defined for \(A = 0^\circ\).
