EXERCISE 8.2
Introduction To Trigonometry • 4 Questions
Question 1
Hint available
Evaluate the following : (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° – sin2 60° (iii) cos 45° sec 30° + cosec 30° (iv) sin 30° + tan 45° – cosec 60° sec 30° + cos 60° + cot 45° (v) 2 2 2 2 2 5 cos 60 4 sec 30 tan 45 sin 30 cos 30
Key Idea
Use the standard trigonometric values for 30°, 45°, 60° and the fundamental identities:
- $
sin(A+B)=sinA\,cosB+cosA\,sinB$
- $
cos^2\theta+sin^2\theta=1$
- Reciprocal definitions $\sec\theta=1/\cos\theta$, $\cosec\theta=1/\sin\theta$, $\cot\theta=1/\tan\theta$.
All calculations are performed by substituting the exact values $\sin30°=\frac12$, $\cos30°=\frac{\sqrt3}{2}$, $\sin45°=\cos45°=\frac{\sqrt2}{2}$, $\sin60°=\cos60°=\frac{\sqrt3}{2}$, $\tan45°=1$, etc.
- $
sin(A+B)=sinA\,cosB+cosA\,sinB$
- $
cos^2\theta+sin^2\theta=1$
- Reciprocal definitions $\sec\theta=1/\cos\theta$, $\cosec\theta=1/\sin\theta$, $\cot\theta=1/\tan\theta$.
All calculations are performed by substituting the exact values $\sin30°=\frac12$, $\cos30°=\frac{\sqrt3}{2}$, $\sin45°=\cos45°=\frac{\sqrt2}{2}$, $\sin60°=\cos60°=\frac{\sqrt3}{2}$, $\tan45°=1$, etc.
Step-by-Step Solution
### (i) $\sin 60^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 60^{\circ}$
Using the identity $\sin(A+B)=\sin A\cos B+\cos A\sin B$,
$$\sin 60^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 60^{\circ}=\sin(60^{\circ}+30^{\circ})=\sin 90^{\circ}=1.$$
Alternatively, substituting the exact values:
$$\sin60^{\circ}=\frac{\sqrt3}{2},\;\cos30^{\circ}=\frac{\sqrt3}{2},\;\sin30^{\circ}=\frac12,\;\cos60^{\circ}=\frac12$$
$$\Rightarrow \frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac12=\frac{3}{4}+\frac{1}{4}=1.$$
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### (ii) $2\tan^{2}45^{\circ}+\cos^{2}30^{\circ}-\sin^{2}60^{\circ}$
\[\tan45^{\circ}=1\Rightarrow\tan^{2}45^{\circ}=1\]
\[\cos30^{\circ}=\frac{\sqrt3}{2}\Rightarrow\cos^{2}30^{\circ}=\frac{3}{4}\]
\[\sin60^{\circ}=\frac{\sqrt3}{2}\Rightarrow\sin^{2}60^{\circ}=\frac{3}{4}\]
Hence
$$2\times1+\frac{3}{4}-\frac{3}{4}=2.$$
---
### (iii) $\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}$
\[\cos45^{\circ}=\frac{\sqrt2}{2},\qquad \sec30^{\circ}=\frac{1}{\cos30^{\circ}}=\frac{2}{\sqrt3}\]
$$\cos45^{\circ}\sec30^{\circ}=\frac{\sqrt2}{2}\cdot\frac{2}{\sqrt3}=\frac{\sqrt2}{\sqrt3}=\frac{\sqrt6}{3}$$
\[\cosec30^{\circ}=\frac{1}{\sin30^{\circ}}=\frac{1}{\frac12}=2\]
Therefore
$$\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}=\frac{\sqrt6}{3}+2=\frac{6+\sqrt6}{3}.$$
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### (iv) $\sin30^{\circ}+\tan45^{\circ}-\cosec60^{\circ}\sec30^{\circ}+\cos60^{\circ}+\cot45^{\circ}$
Values:
\[\sin30^{\circ}=\frac12,\;\tan45^{\circ}=1,\;\cosec60^{\circ}=\frac{2}{\sqrt3},\;\sec30^{\circ}=\frac{2}{\sqrt3},\;\cos60^{\circ}=\frac12,\;\cot45^{\circ}=1\]
Compute the product:
$$\cosec60^{\circ}\sec30^{\circ}=\frac{2}{\sqrt3}\cdot\frac{2}{\sqrt3}=\frac{4}{3}$$
Now sum all terms:
$$\frac12+1-\frac{4}{3}+\frac12+1=\left(\frac12+\frac12\right)+(1+1)-\frac{4}{3}=1+2-\frac{4}{3}=\frac{5}{3}.$$
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### (v) $2\cos60^{\circ}+2\sec30^{\circ}-2\tan45^{\circ}+2\sin30^{\circ}+2\cos30^{\circ}$
Substituting the standard values:
\[2\cos60^{\circ}=2\times\frac12=1\]
\[2\sec30^{\circ}=2\times\frac{2}{\sqrt3}=\frac{4}{\sqrt3}=\frac{4\sqrt3}{3}\]
\[2\tan45^{\circ}=2\times1=2\]
\[2\sin30^{\circ}=2\times\frac12=1\]
\[2\cos30^{\circ}=2\times\frac{\sqrt3}{2}=\sqrt3\]
Adding them:
$$1+\frac{4\sqrt3}{3}-2+1+\sqrt3=\left(1+1-2\right)+\left(\frac{4\sqrt3}{3}+\sqrt3\right)=0+\frac{4\sqrt3+3\sqrt3}{3}=\frac{7\sqrt3}{3}.$$
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Final Answers
- (i) $1$
- (ii) $2$
- (iii) $\dfrac{6+\sqrt6}{3}$
- (iv) $\dfrac{5}{3}$
- (v) $\dfrac{7\sqrt3}{3}$
Using the identity $\sin(A+B)=\sin A\cos B+\cos A\sin B$,
$$\sin 60^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 60^{\circ}=\sin(60^{\circ}+30^{\circ})=\sin 90^{\circ}=1.$$
Alternatively, substituting the exact values:
$$\sin60^{\circ}=\frac{\sqrt3}{2},\;\cos30^{\circ}=\frac{\sqrt3}{2},\;\sin30^{\circ}=\frac12,\;\cos60^{\circ}=\frac12$$
$$\Rightarrow \frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac12=\frac{3}{4}+\frac{1}{4}=1.$$
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### (ii) $2\tan^{2}45^{\circ}+\cos^{2}30^{\circ}-\sin^{2}60^{\circ}$
\[\tan45^{\circ}=1\Rightarrow\tan^{2}45^{\circ}=1\]
\[\cos30^{\circ}=\frac{\sqrt3}{2}\Rightarrow\cos^{2}30^{\circ}=\frac{3}{4}\]
\[\sin60^{\circ}=\frac{\sqrt3}{2}\Rightarrow\sin^{2}60^{\circ}=\frac{3}{4}\]
Hence
$$2\times1+\frac{3}{4}-\frac{3}{4}=2.$$
---
### (iii) $\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}$
\[\cos45^{\circ}=\frac{\sqrt2}{2},\qquad \sec30^{\circ}=\frac{1}{\cos30^{\circ}}=\frac{2}{\sqrt3}\]
$$\cos45^{\circ}\sec30^{\circ}=\frac{\sqrt2}{2}\cdot\frac{2}{\sqrt3}=\frac{\sqrt2}{\sqrt3}=\frac{\sqrt6}{3}$$
\[\cosec30^{\circ}=\frac{1}{\sin30^{\circ}}=\frac{1}{\frac12}=2\]
Therefore
$$\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}=\frac{\sqrt6}{3}+2=\frac{6+\sqrt6}{3}.$$
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### (iv) $\sin30^{\circ}+\tan45^{\circ}-\cosec60^{\circ}\sec30^{\circ}+\cos60^{\circ}+\cot45^{\circ}$
Values:
\[\sin30^{\circ}=\frac12,\;\tan45^{\circ}=1,\;\cosec60^{\circ}=\frac{2}{\sqrt3},\;\sec30^{\circ}=\frac{2}{\sqrt3},\;\cos60^{\circ}=\frac12,\;\cot45^{\circ}=1\]
Compute the product:
$$\cosec60^{\circ}\sec30^{\circ}=\frac{2}{\sqrt3}\cdot\frac{2}{\sqrt3}=\frac{4}{3}$$
Now sum all terms:
$$\frac12+1-\frac{4}{3}+\frac12+1=\left(\frac12+\frac12\right)+(1+1)-\frac{4}{3}=1+2-\frac{4}{3}=\frac{5}{3}.$$
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### (v) $2\cos60^{\circ}+2\sec30^{\circ}-2\tan45^{\circ}+2\sin30^{\circ}+2\cos30^{\circ}$
Substituting the standard values:
\[2\cos60^{\circ}=2\times\frac12=1\]
\[2\sec30^{\circ}=2\times\frac{2}{\sqrt3}=\frac{4}{\sqrt3}=\frac{4\sqrt3}{3}\]
\[2\tan45^{\circ}=2\times1=2\]
\[2\sin30^{\circ}=2\times\frac12=1\]
\[2\cos30^{\circ}=2\times\frac{\sqrt3}{2}=\sqrt3\]
Adding them:
$$1+\frac{4\sqrt3}{3}-2+1+\sqrt3=\left(1+1-2\right)+\left(\frac{4\sqrt3}{3}+\sqrt3\right)=0+\frac{4\sqrt3+3\sqrt3}{3}=\frac{7\sqrt3}{3}.$$
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Final Answers
- (i) $1$
- (ii) $2$
- (iii) $\dfrac{6+\sqrt6}{3}$
- (iv) $\dfrac{5}{3}$
- (v) $\dfrac{7\sqrt3}{3}$
Question 2
Hint available
Choose the correct option and justify your choice : (i) 2 2 tan 30 1 tan 30 (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (ii) 2 2 1 tan 45 1 tan 45 (A) tan 90° (B) 1 (C) sin 45° (D) 0 (iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60° (iv) 2 2 tan 30 1 tan 30 (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Key Idea
Use the exact values of trigonometric ratios for 30° and 45°, the double‑angle identity \(\sin 2A = 2\sin A\cos A\), and simple algebraic manipulation. For any angle \(\theta\), \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\tan 45^{\circ}=1\). Also \(\tan 60^{\circ}=\sqrt{3}\). The identity \(\sin 2A = 2\sin A\cos A\) reduces the equation \(\sin 2A = 2\sin A\) to \(\cos A = 1\) (or \(\sin A =0\)).
Step-by-Step Solution
### (i) \(2\theta = 2\tan30^{\circ}+\tan30^{\circ}\)
1. \(\tan30^{\circ}=\dfrac{1}{\sqrt{3}}\).
2. \(2\tan30^{\circ}+\tan30^{\circ}=3\times\dfrac{1}{\sqrt{3}}=\sqrt{3}\).
3. \(\sqrt{3}=\tan60^{\circ}\) (since \(\tan60^{\circ}=\sqrt{3}\)).
Answer: (C) \(\tan60^{\circ}\).
### (ii) \(2\theta = 2\tan45^{\circ}-\tan45^{\circ}\)
1. \(\tan45^{\circ}=1\).
2. \(2\tan45^{\circ}-\tan45^{\circ}=2\times1-1=1\).
3. The value obtained is the number 1.
Answer: (B) 1.
### (iii) Solve \(\sin 2A = 2\sin A\)
1. Use the double‑angle identity: \(\sin 2A = 2\sin A\cos A\).
2. Set \(2\sin A\cos A = 2\sin A\).
3. Divide both sides by \(2\sin A\) (possible when \(\sin A
eq 0\)):
\[\cos A = 1\].
4. \(\cos A = 1\) gives \(A = 0^{\circ}\) (within the principal range 0°–90°).
5. If \(\sin A = 0\), then \(A = 0^{\circ}\) as well.
Answer: (A) \(0^{\circ}\).
### (iv) \(2\theta = \dfrac{2}{\tan30^{\circ}}-\dfrac{1}{\tan30^{\circ}}\)
1. \(\tan30^{\circ}=\dfrac{1}{\sqrt{3}}\).
2. Compute the reciprocals: \(\dfrac{1}{\tan30^{\circ}} = \sqrt{3}\).
3. \(\dfrac{2}{\tan30^{\circ}}-\dfrac{1}{\tan30^{\circ}} = 2\sqrt{3}-\sqrt{3}=\sqrt{3}\).
4. \(\sqrt{3}=\tan60^{\circ}\).
Answer: (C) \(\tan60^{\circ}\).
1. \(\tan30^{\circ}=\dfrac{1}{\sqrt{3}}\).
2. \(2\tan30^{\circ}+\tan30^{\circ}=3\times\dfrac{1}{\sqrt{3}}=\sqrt{3}\).
3. \(\sqrt{3}=\tan60^{\circ}\) (since \(\tan60^{\circ}=\sqrt{3}\)).
Answer: (C) \(\tan60^{\circ}\).
### (ii) \(2\theta = 2\tan45^{\circ}-\tan45^{\circ}\)
1. \(\tan45^{\circ}=1\).
2. \(2\tan45^{\circ}-\tan45^{\circ}=2\times1-1=1\).
3. The value obtained is the number 1.
Answer: (B) 1.
### (iii) Solve \(\sin 2A = 2\sin A\)
1. Use the double‑angle identity: \(\sin 2A = 2\sin A\cos A\).
2. Set \(2\sin A\cos A = 2\sin A\).
3. Divide both sides by \(2\sin A\) (possible when \(\sin A
eq 0\)):
\[\cos A = 1\].
4. \(\cos A = 1\) gives \(A = 0^{\circ}\) (within the principal range 0°–90°).
5. If \(\sin A = 0\), then \(A = 0^{\circ}\) as well.
Answer: (A) \(0^{\circ}\).
### (iv) \(2\theta = \dfrac{2}{\tan30^{\circ}}-\dfrac{1}{\tan30^{\circ}}\)
1. \(\tan30^{\circ}=\dfrac{1}{\sqrt{3}}\).
2. Compute the reciprocals: \(\dfrac{1}{\tan30^{\circ}} = \sqrt{3}\).
3. \(\dfrac{2}{\tan30^{\circ}}-\dfrac{1}{\tan30^{\circ}} = 2\sqrt{3}-\sqrt{3}=\sqrt{3}\).
4. \(\sqrt{3}=\tan60^{\circ}\).
Answer: (C) \(\tan60^{\circ}\).
Question 3
Hint available
If tan (A + B) = 3 and tan (A – B) = 1 3 ; 0° < A + B 90°; A > B, find A and B.
Key Idea
Use the tan addition and subtraction formulas: \(\tan(A\pm B)=\dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\). Set \(x=\tan A\) and \(y=\tan B\) and solve the resulting system of two equations for \(x\) and \(y\). Then obtain the angles from their tangents, keeping in mind the given range of the angles.
Step-by-Step Solution
1. Let \(x=\tan A\) and \(y=\tan B\).\
2. Using the addition formula: \[\tan(A+B)=\frac{x+y}{1-xy}=3\] ⇒ \(x+y=3(1-xy)=3-3xy\) …(i)\
3. Using the subtraction formula: \[\tan(A-B)=\frac{x-y}{1+xy}=\frac13\] ⇒ \(3(x-y)=1+xy\) ⇒ \(3x-3y-xy=1\) …(ii)\
4. Rewrite (i) as \(x+y+3xy=3\).\
5. Multiply (i) by 3: \(3x+3y+9xy=9\).\
6. Add this to (ii): \( (3x+3y+9xy)+(3x-3y-xy)=9+1\) ⇒ \(6x+8xy=10\) ⇒ \(x(3+4y)=5\) ⇒ \(x=\frac{5}{3+4y}\) …(iii)\
7. Substitute (iii) into (i): \[\frac{5}{3+4y}+y+3y\frac{5}{3+4y}=3\]\
8. Combine the fractions: \[\frac{5+15y}{3+4y}+y=3\] ⇒ \[\frac{5(1+3y)}{3+4y}+y=3\]\
9. Multiply by \(3+4y\): \[5(1+3y)+y(3+4y)=3(3+4y)\]\
10. Expand: \[5+15y+3y+4y^{2}=9+12y\] ⇒ \[4y^{2}+18y+5=9+12y\] ⇒ \[4y^{2}+6y-4=0\]\
11. Divide by 2: \[2y^{2}+3y-2=0\]. Solve the quadratic: \(D=3^{2}-4\cdot2(-2)=9+16=25\).\
12. \[y=\frac{-3\pm5}{4}\] ⇒ \(y=\frac{2}{4}=\frac12\) or \(y=\frac{-8}{4}=-2\). Since \(0°0\). Hence \(y=\tan B=\frac12\).\
13. From (iii): \[x=\frac{5}{3+4y}=\frac{5}{3+4\cdot\frac12}=\frac{5}{3+2}=\frac{5}{5}=1\] ⇒ \(\tan A = x = 1\).\
14. Therefore \(A=\tan^{-1}(1)=45^{\circ}\).\
15. \(B=\tan^{-1}\left(\frac12\right)\). In degrees, \(B\approx 26.6^{\circ}\).\
16. Check: \(A+B\approx71.6^{\circ}\) (within the given range) and \(A-B\approx18.4^{\circ}\) with \(\tan(A+B)=3\) and \(\tan(A-B)=\frac13\). The solution satisfies all conditions.
2. Using the addition formula: \[\tan(A+B)=\frac{x+y}{1-xy}=3\] ⇒ \(x+y=3(1-xy)=3-3xy\) …(i)\
3. Using the subtraction formula: \[\tan(A-B)=\frac{x-y}{1+xy}=\frac13\] ⇒ \(3(x-y)=1+xy\) ⇒ \(3x-3y-xy=1\) …(ii)\
4. Rewrite (i) as \(x+y+3xy=3\).\
5. Multiply (i) by 3: \(3x+3y+9xy=9\).\
6. Add this to (ii): \( (3x+3y+9xy)+(3x-3y-xy)=9+1\) ⇒ \(6x+8xy=10\) ⇒ \(x(3+4y)=5\) ⇒ \(x=\frac{5}{3+4y}\) …(iii)\
7. Substitute (iii) into (i): \[\frac{5}{3+4y}+y+3y\frac{5}{3+4y}=3\]\
8. Combine the fractions: \[\frac{5+15y}{3+4y}+y=3\] ⇒ \[\frac{5(1+3y)}{3+4y}+y=3\]\
9. Multiply by \(3+4y\): \[5(1+3y)+y(3+4y)=3(3+4y)\]\
10. Expand: \[5+15y+3y+4y^{2}=9+12y\] ⇒ \[4y^{2}+18y+5=9+12y\] ⇒ \[4y^{2}+6y-4=0\]\
11. Divide by 2: \[2y^{2}+3y-2=0\]. Solve the quadratic: \(D=3^{2}-4\cdot2(-2)=9+16=25\).\
12. \[y=\frac{-3\pm5}{4}\] ⇒ \(y=\frac{2}{4}=\frac12\) or \(y=\frac{-8}{4}=-2\). Since \(0°
13. From (iii): \[x=\frac{5}{3+4y}=\frac{5}{3+4\cdot\frac12}=\frac{5}{3+2}=\frac{5}{5}=1\] ⇒ \(\tan A = x = 1\).\
14. Therefore \(A=\tan^{-1}(1)=45^{\circ}\).\
15. \(B=\tan^{-1}\left(\frac12\right)\). In degrees, \(B\approx 26.6^{\circ}\).\
16. Check: \(A+B\approx71.6^{\circ}\) (within the given range) and \(A-B\approx18.4^{\circ}\) with \(\tan(A+B)=3\) and \(\tan(A-B)=\frac13\). The solution satisfies all conditions.
Question 4
Hint available
State whether the following are true or false. Justify your answer. (i) sin (A + B) = sin A + sin B. (ii) The value of sin increases as increases. (iii) The value of cos increases as increases. (iv) sin = cos for all values of . (v) cot A is not defined for A = 0°. 128
Key Idea
Use the fundamental trigonometric identities and the monotonic behaviour of sine and cosine in the first quadrant (0° ≤ θ ≤ 90°). Recall that sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, and cot θ = cos θ / sin θ.
Step-by-Step Solution
1. (i) sin(A+B) = sin A + sin B
- The correct addition formula (proved in the textbook) is
$$\sin(A+B)=\sin A\cos B+\cos A\sin B.$$
- Since $\cos B$ and $\cos A$ are generally not equal to 1, the right‑hand side is not equal to $\sin A+\sin B$.
- Counter‑example: Take $A=30^{\circ}, B=60^{\circ}$. Then $\sin(90^{\circ})=1$, whereas $\sin30^{\circ}+\sin60^{\circ}=0.5+0.866=1.366
eq1$.
- Hence the statement is False.
2. (ii) The value of sin θ increases as θ increases
- From the unit‑circle definition, for $0^{\circ}\le θ\le 90^{\circ}$ the ordinate of the point on the circle increases with the angle; therefore $\sin θ$ is a strictly increasing function in this interval.
- In the textbook it is mentioned that "the sine of an acute angle increases as the angle increases".
- Hence, within the principal range $0^{\circ}\le θ\le 90^{\circ}$ the statement is True (it is false outside this range, but the question is understood in the context of acute angles).
3. (iii) The value of cos θ increases as θ increases
- In the same interval $0^{\circ}\le θ\le 90^{\circ}$ the abscissa of the point on the unit circle decreases as the angle increases; consequently $\cos θ$ is a decreasing function.
- Example: $\cos30^{\circ}=\frac{\sqrt3}{2}\approx0.866$, while $\cos60^{\circ}=\frac12=0.5$.
- Therefore the statement is False.
4. (iv) sin θ = cos θ for all values of θ
- Equality holds only when $\tan θ = 1$, i.e. $θ = 45^{\circ} + n\times180^{\circ}$ (or $θ = \frac{\pi}{4}+n\pi$ in radians).
- For most angles the two functions have different values (e.g., $\sin30^{\circ}=0.5$, $\cos30^{\circ}=0.866$).
- Hence the statement is False.
5. (v) cot A is not defined for A = 0°
- By definition $\cot A = \frac{\cos A}{\sin A}$.
- At $A=0^{\circ}$, $\sin0^{\circ}=0$, so the denominator becomes zero and the expression is undefined.
- Therefore the statement is True.
Summary of answers:
(i) False, (ii) True, (iii) False, (iv) False, (v) True.
- The correct addition formula (proved in the textbook) is
$$\sin(A+B)=\sin A\cos B+\cos A\sin B.$$
- Since $\cos B$ and $\cos A$ are generally not equal to 1, the right‑hand side is not equal to $\sin A+\sin B$.
- Counter‑example: Take $A=30^{\circ}, B=60^{\circ}$. Then $\sin(90^{\circ})=1$, whereas $\sin30^{\circ}+\sin60^{\circ}=0.5+0.866=1.366
eq1$.
- Hence the statement is False.
2. (ii) The value of sin θ increases as θ increases
- From the unit‑circle definition, for $0^{\circ}\le θ\le 90^{\circ}$ the ordinate of the point on the circle increases with the angle; therefore $\sin θ$ is a strictly increasing function in this interval.
- In the textbook it is mentioned that "the sine of an acute angle increases as the angle increases".
- Hence, within the principal range $0^{\circ}\le θ\le 90^{\circ}$ the statement is True (it is false outside this range, but the question is understood in the context of acute angles).
3. (iii) The value of cos θ increases as θ increases
- In the same interval $0^{\circ}\le θ\le 90^{\circ}$ the abscissa of the point on the unit circle decreases as the angle increases; consequently $\cos θ$ is a decreasing function.
- Example: $\cos30^{\circ}=\frac{\sqrt3}{2}\approx0.866$, while $\cos60^{\circ}=\frac12=0.5$.
- Therefore the statement is False.
4. (iv) sin θ = cos θ for all values of θ
- Equality holds only when $\tan θ = 1$, i.e. $θ = 45^{\circ} + n\times180^{\circ}$ (or $θ = \frac{\pi}{4}+n\pi$ in radians).
- For most angles the two functions have different values (e.g., $\sin30^{\circ}=0.5$, $\cos30^{\circ}=0.866$).
- Hence the statement is False.
5. (v) cot A is not defined for A = 0°
- By definition $\cot A = \frac{\cos A}{\sin A}$.
- At $A=0^{\circ}$, $\sin0^{\circ}=0$, so the denominator becomes zero and the expression is undefined.
- Therefore the statement is True.
Summary of answers:
(i) False, (ii) True, (iii) False, (iv) False, (v) True.