EXAMPLES
Introduction To Trigonometry • 12 Questions
Question 1
Hint available
Given tan A = 4 3 , find the other trigonometric ratios of the angle A.
Key Idea
Use the definition of tangent in a right‑angled triangle (tan A = opposite/adjacent). Choose convenient integer lengths for the opposite and adjacent sides, apply the Pythagorean theorem to obtain the hypotenuse, and then use the reciprocal relations to obtain the remaining trigonometric ratios.
Step-by-Step Solution
1. From the definition, \(\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{3}\).
Choose a right‑angled triangle with opposite side = 4 units and adjacent side = 3 units.\
2. Compute the hypotenuse using Pythagoras: \[\text{hypotenuse} = \sqrt{4^{2}+3^{2}} = \sqrt{16+9}=\sqrt{25}=5\]
3. Find \(\sin A\) and \(\cos A\):\
\[\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}, \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{5}\]
4. Obtain the remaining ratios using reciprocal identities:\
\[\cot A = \frac{1}{\tan A}=\frac{3}{4},\qquad \sec A = \frac{1}{\cos A}=\frac{5}{3},\qquad \csc A = \frac{1}{\sin A}=\frac{5}{4}\]
5. Summarise all the trigonometric ratios of \(A\):\
\[\tan A = \frac{4}{3},\; \cot A = \frac{3}{4},\; \sin A = \frac{4}{5},\; \cos A = \frac{3}{5},\; \sec A = \frac{5}{3},\; \csc A = \frac{5}{4}\]
Choose a right‑angled triangle with opposite side = 4 units and adjacent side = 3 units.\
2. Compute the hypotenuse using Pythagoras: \[\text{hypotenuse} = \sqrt{4^{2}+3^{2}} = \sqrt{16+9}=\sqrt{25}=5\]
3. Find \(\sin A\) and \(\cos A\):\
\[\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}, \qquad \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{5}\]
4. Obtain the remaining ratios using reciprocal identities:\
\[\cot A = \frac{1}{\tan A}=\frac{3}{4},\qquad \sec A = \frac{1}{\cos A}=\frac{5}{3},\qquad \csc A = \frac{1}{\sin A}=\frac{5}{4}\]
5. Summarise all the trigonometric ratios of \(A\):\
\[\tan A = \frac{4}{3},\; \cot A = \frac{3}{4},\; \sin A = \frac{4}{5},\; \cos A = \frac{3}{5},\; \sec A = \frac{5}{3},\; \csc A = \frac{5}{4}\]
Question 2
Hint available
If B and Q are acute angles such that sin B = sin Q, then prove that B = Q.
Key Idea
For acute angles (0° < θ < 90°) the sine function is strictly increasing; alternatively, use the identity \(\sin C-\sin D = 2\cos\frac{C+D}{2}\sin\frac{C-D}{2}\) and the fact that \(\cos\) of an acute angle is never zero.
Step-by-Step Solution
1. Given \(\sin B = \sin Q\) with \(0^{\circ}
2. Subtract the two sides:
$$\sin B - \sin Q = 0.$$
3. Use the trigonometric identity:
$$\sin B - \sin Q = 2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}.$$
Hence
$$2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}=0.$$
4. For acute angles, \(\frac{B+Q}{2}\) is also acute (since the average of two acute angles is acute). Therefore \(\cos\frac{B+Q}{2}
eq 0\).
5. The product being zero forces the second factor to be zero:
$$\sin\frac{B-Q}{2}=0.$$
6. The sine of an angle is zero only when the angle is an integer multiple of \(180^{\circ}\). Because \(\frac{B-Q}{2}\) lies between \(-45^{\circ}\) and \(45^{\circ}\) (both acute), the only possible multiple is \(0^{\circ}\).
Hence
$$\frac{B-Q}{2}=0^{\circ}\quad\Rightarrow\quad B-Q=0^{\circ}.$$
7. Therefore, \(B = Q\).
8. Conclusion: If two acute angles have equal sines, the angles themselves are equal.
*Alternative reasoning (monotonicity):* Since \(\sin\theta\) is strictly increasing on \(0^{\circ}<\theta<90^{\circ}\), equality of sines directly implies equality of the angles.
2. Subtract the two sides:
$$\sin B - \sin Q = 0.$$
3. Use the trigonometric identity:
$$\sin B - \sin Q = 2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}.$$
Hence
$$2\cos\frac{B+Q}{2}\,\sin\frac{B-Q}{2}=0.$$
4. For acute angles, \(\frac{B+Q}{2}\) is also acute (since the average of two acute angles is acute). Therefore \(\cos\frac{B+Q}{2}
eq 0\).
5. The product being zero forces the second factor to be zero:
$$\sin\frac{B-Q}{2}=0.$$
6. The sine of an angle is zero only when the angle is an integer multiple of \(180^{\circ}\). Because \(\frac{B-Q}{2}\) lies between \(-45^{\circ}\) and \(45^{\circ}\) (both acute), the only possible multiple is \(0^{\circ}\).
Hence
$$\frac{B-Q}{2}=0^{\circ}\quad\Rightarrow\quad B-Q=0^{\circ}.$$
7. Therefore, \(B = Q\).
8. Conclusion: If two acute angles have equal sines, the angles themselves are equal.
*Alternative reasoning (monotonicity):* Since \(\sin\theta\) is strictly increasing on \(0^{\circ}<\theta<90^{\circ}\), equality of sines directly implies equality of the angles.
Question 3
Hint available
Consider ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ABC = (see Fig. 8.10). Determine the values of (i) cos2 + sin2 , (ii) cos2 – sin2
Key Idea
Use the Pythagorean theorem to find the missing side of the right‑angled triangle, then apply the definitions of sine and cosine for the acute angle θ. Finally employ the fundamental trigonometric identity $\sin^2\theta+\cos^2\theta=1$ and the difference formula $\cos^2\theta-\sin^2\theta=\cos2\theta$.
Step-by-Step Solution
1. Find the third side AC using Pythagoras:
$$AB^2 = AC^2 + BC^2 \Rightarrow AC^2 = 29^2 - 21^2 = 841 - 441 = 400$$
$$\therefore\; AC = \sqrt{400}=20\text{ units}$$
2. Express sinθ and cosθ (θ is at vertex B):
- Opposite side to θ = AC = 20
- Adjacent side to θ = BC = 21
- Hypotenuse = AB = 29
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{20}{29}$$
$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{21}{29}$$
3. Compute $\cos^2\theta + \sin^2\theta$:
$$\cos^2\theta + \sin^2\theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2
= \frac{441 + 400}{29^2}
= \frac{841}{841}
= 1$$
4. Compute $\cos^2\theta - \sin^2\theta$:
$$\cos^2\theta - \sin^2\theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2
= \frac{441 - 400}{29^2}
= \frac{41}{841}$$
(This is also $\cos2\theta$.)
Thus the required values are obtained.
$$AB^2 = AC^2 + BC^2 \Rightarrow AC^2 = 29^2 - 21^2 = 841 - 441 = 400$$
$$\therefore\; AC = \sqrt{400}=20\text{ units}$$
2. Express sinθ and cosθ (θ is at vertex B):
- Opposite side to θ = AC = 20
- Adjacent side to θ = BC = 21
- Hypotenuse = AB = 29
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{20}{29}$$
$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{21}{29}$$
3. Compute $\cos^2\theta + \sin^2\theta$:
$$\cos^2\theta + \sin^2\theta = \left(\frac{21}{29}\right)^2 + \left(\frac{20}{29}\right)^2
= \frac{441 + 400}{29^2}
= \frac{841}{841}
= 1$$
4. Compute $\cos^2\theta - \sin^2\theta$:
$$\cos^2\theta - \sin^2\theta = \left(\frac{21}{29}\right)^2 - \left(\frac{20}{29}\right)^2
= \frac{441 - 400}{29^2}
= \frac{41}{841}$$
(This is also $\cos2\theta$.)
Thus the required values are obtained.
Question 4
Hint available
In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Key Idea
Use the definition of tangent in a right triangle (tan A = opposite/adjacent = sin A / cos A) and the Pythagorean identity \(\sin^2 A + \cos^2 A = 1\). From tan A = 1 we get \(\sin A = \cos A\); substituting into the identity gives the exact values of \(\sin A\) and \(\cos A\), which can then be used to evaluate \(2\sin A\cos A\).
Step-by-Step Solution
1. In \(\triangle ABC\) right‑angled at \(B\), the trigonometric ratios are defined as:
$$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin A}{\cos A}.$$
2. Given \(\tan A = 1\), we have
$$\frac{\sin A}{\cos A} = 1 \quad\Rightarrow\quad \sin A = \cos A. $$
3. Use the fundamental identity for any angle:
$$\sin^2 A + \cos^2 A = 1.$$
4. Substitute \(\sin A = \cos A\) into the identity:
$$\sin^2 A + \sin^2 A = 1 \;\Rightarrow\; 2\sin^2 A = 1 \;\Rightarrow\; \sin^2 A = \frac{1}{2}.$$
5. Hence
$$\sin A = \cos A = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$
6. Now evaluate the expression to be verified:
$$2\sin A \cos A = 2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)
= 2\times \frac{2}{4}
= 1.$$
7. Therefore, the required identity is verified:
$$2\sin A \cos A = 1.$$
$$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin A}{\cos A}.$$
2. Given \(\tan A = 1\), we have
$$\frac{\sin A}{\cos A} = 1 \quad\Rightarrow\quad \sin A = \cos A. $$
3. Use the fundamental identity for any angle:
$$\sin^2 A + \cos^2 A = 1.$$
4. Substitute \(\sin A = \cos A\) into the identity:
$$\sin^2 A + \sin^2 A = 1 \;\Rightarrow\; 2\sin^2 A = 1 \;\Rightarrow\; \sin^2 A = \frac{1}{2}.$$
5. Hence
$$\sin A = \cos A = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$
6. Now evaluate the expression to be verified:
$$2\sin A \cos A = 2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)
= 2\times \frac{2}{4}
= 1.$$
7. Therefore, the required identity is verified:
$$2\sin A \cos A = 1.$$
Question 5
Hint available
In OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q.
Key Idea
Use Pythagoras theorem to find the unknown sides of the right‑angled triangle and then apply the definitions of sine and cosine for the acute angle Q: \(\sin Q = \frac{\text{opposite side}}{\text{hypotenuse}}\) and \(\cos Q = \frac{\text{adjacent side}}{\text{hypotenuse}}\).
Step-by-Step Solution
1. Let \(PQ = x\) cm. Then \(OQ = x + 1\) cm (given).\
2. Since the triangle is right‑angled at P, apply Pythagoras theorem:
$$OP^{2} + PQ^{2} = OQ^{2}$$
Substituting the known values:
$$7^{2} + x^{2} = (x+1)^{2}$$
3. Expand and simplify:
$$49 + x^{2} = x^{2} + 2x + 1$$
$$49 = 2x + 1$$
$$2x = 48$$
$$x = 24$$
Hence, \(PQ = 24\) cm and \(OQ = 24 + 1 = 25\) cm.
4. Identify the sides with respect to angle \(Q\):
- Opposite side to \(Q\) = \(OP = 7\) cm.
- Adjacent side to \(Q\) = \(PQ = 24\) cm.
- Hypotenuse = \(OQ = 25\) cm.
5. Use the definitions:
$$\sin Q = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$$
$$\cos Q = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$$
2. Since the triangle is right‑angled at P, apply Pythagoras theorem:
$$OP^{2} + PQ^{2} = OQ^{2}$$
Substituting the known values:
$$7^{2} + x^{2} = (x+1)^{2}$$
3. Expand and simplify:
$$49 + x^{2} = x^{2} + 2x + 1$$
$$49 = 2x + 1$$
$$2x = 48$$
$$x = 24$$
Hence, \(PQ = 24\) cm and \(OQ = 24 + 1 = 25\) cm.
4. Identify the sides with respect to angle \(Q\):
- Opposite side to \(Q\) = \(OP = 7\) cm.
- Adjacent side to \(Q\) = \(PQ = 24\) cm.
- Hypotenuse = \(OQ = 25\) cm.
5. Use the definitions:
$$\sin Q = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$$
$$\cos Q = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$$
Question 6
Hint available
In ABC, right-angled at B, AB = 5 cm and ACB = 30° (see Fig. 8.19). Determine the lengths of the sides BC and AC.
Key Idea
Use the trigonometric ratios for a right‑angled triangle: \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\theta = 30^{\circ}\) we have \(\sin 30^{\circ}=\frac{1}{2}\) and \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\). The given side AB is opposite \(30^{\circ}\).
Step-by-Step Solution
1. Identify the known parts of the triangle:\\
- \(\angle B = 90^{\circ}\) (right angle)
- \(\angle C = 30^{\circ}\)
- \(\angle A = 180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}\)
- Side \(AB = 5\,\text{cm}\) is opposite \(\angle C\).\
2. Apply the definition of sine for \(\angle C\):
$$\sin 30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC}$$\
Since \(\sin 30^{\circ}=\frac{1}{2}\),
$$\frac{1}{2}=\frac{5}{AC}\quad\Rightarrow\quad AC = 10\,\text{cm}.$$\
3. Apply the definition of cosine for \(\angle C\):
$$\cos 30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AC}$$\
With \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\) and \(AC=10\) cm,
$$\frac{\sqrt{3}}{2}=\frac{BC}{10}\quad\Rightarrow\quad BC = 10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\,\text{cm}.$$\
4. Hence the required lengths are:
- \(BC = 5\sqrt{3}\,\text{cm}\)
- \(AC = 10\,\text{cm}\).
- \(\angle B = 90^{\circ}\) (right angle)
- \(\angle C = 30^{\circ}\)
- \(\angle A = 180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}\)
- Side \(AB = 5\,\text{cm}\) is opposite \(\angle C\).\
2. Apply the definition of sine for \(\angle C\):
$$\sin 30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{AC}$$\
Since \(\sin 30^{\circ}=\frac{1}{2}\),
$$\frac{1}{2}=\frac{5}{AC}\quad\Rightarrow\quad AC = 10\,\text{cm}.$$\
3. Apply the definition of cosine for \(\angle C\):
$$\cos 30^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BC}{AC}$$\
With \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}\) and \(AC=10\) cm,
$$\frac{\sqrt{3}}{2}=\frac{BC}{10}\quad\Rightarrow\quad BC = 10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\,\text{cm}.$$\
4. Hence the required lengths are:
- \(BC = 5\sqrt{3}\,\text{cm}\)
- \(AC = 10\,\text{cm}\).
Question 7
Hint available
In PQR, right-angled at Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. Determine QPR and PRQ.
Key Idea
In a right‑angled triangle, the trigonometric ratios (sin, cos, tan) relate the lengths of the sides to the acute angles. Since the side adjacent to ∠QPR (PQ) and the hypotenuse (PR) are known, we can use the cosine ratio to find ∠QPR, and then use the fact that the two acute angles of a right triangle are complementary.
Step-by-Step Solution
1. Identify the known sides\
- Right angle at Q ⇒ PR is the hypotenuse.\
- Given: \(PQ = 3\text{ cm}\) (adjacent to \(\angle QPR\)), \(PR = 6\text{ cm}\) (hypotenuse).\
2. Use the cosine definition for \(\angle QPR\):\
$$\cos \angle QPR = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}.$$\
3. Find the angle whose cosine is \(\frac12\):\
$$\angle QPR = \cos^{-1}\left(\frac12\right) = 60^{\circ}.$$\
4. Determine the other acute angle \(\angle PRQ\) using the complementary property of a right triangle (the two acute angles sum to \(90^{\circ}\)):\
$$\angle PRQ = 90^{\circ} - \angle QPR = 90^{\circ} - 60^{\circ} = 30^{\circ}.$$\
5. (Optional) Find the remaining side QR for verification using the sine ratio:\
$$\sin 60^{\circ} = \frac{QR}{PR} \;\Rightarrow\; QR = PR \cdot \sin 60^{\circ} = 6 \times \frac{\sqrt3}{2} = 3\sqrt3\text{ cm}.$$\
Thus, \(\angle QPR = 60^{\circ}\) and \(\angle PRQ = 30^{\circ}\).
- Right angle at Q ⇒ PR is the hypotenuse.\
- Given: \(PQ = 3\text{ cm}\) (adjacent to \(\angle QPR\)), \(PR = 6\text{ cm}\) (hypotenuse).\
2. Use the cosine definition for \(\angle QPR\):\
$$\cos \angle QPR = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}.$$\
3. Find the angle whose cosine is \(\frac12\):\
$$\angle QPR = \cos^{-1}\left(\frac12\right) = 60^{\circ}.$$\
4. Determine the other acute angle \(\angle PRQ\) using the complementary property of a right triangle (the two acute angles sum to \(90^{\circ}\)):\
$$\angle PRQ = 90^{\circ} - \angle QPR = 90^{\circ} - 60^{\circ} = 30^{\circ}.$$\
5. (Optional) Find the remaining side QR for verification using the sine ratio:\
$$\sin 60^{\circ} = \frac{QR}{PR} \;\Rightarrow\; QR = PR \cdot \sin 60^{\circ} = 6 \times \frac{\sqrt3}{2} = 3\sqrt3\text{ cm}.$$\
Thus, \(\angle QPR = 60^{\circ}\) and \(\angle PRQ = 30^{\circ}\).
Question 8
Hint available
If sin (A – B) = 1 , 2 cos (A + B) = 1 , 2 0° < A + B 90°, A > B, find A and B.
Key Idea
Use the known values of sine and cosine for special angles (90°, 60°) and solve the resulting simultaneous linear equations in A and B.
Step-by-Step Solution
1. From \(\sin(A-B)=1\) we know that \(A-B = 90^{\circ}\) because \(\sin\theta = 1\) only when \(\theta = 90^{\circ}+360^{\circ}k\). Since \(A>B\) the smallest positive value is taken: \(A-B = 90^{\circ}\).\
2. From \(2\cos(A+B)=1\) we get \(\cos(A+B)=\frac12\). In the interval \(0^{\circ}3. Now we have a system of two linear equations:\
\[\begin{cases} A - B = 90^{\circ} \\ A + B = 60^{\circ} \end{cases}\]\
4. Add the two equations: \(2A = 150^{\circ}\) \(\Rightarrow\) \(A = 75^{\circ}\).\
5. Substitute \(A\) in \(A + B = 60^{\circ}\): \(75^{\circ}+B = 60^{\circ}\) \(\Rightarrow\) \(B = -15^{\circ}\).\
6. Verify the conditions: \(A>B\) (75° > -15°) and \(0^{\circ}Thus the required angles are \(A = 75^{\circ}\) and \(B = -15^{\circ}\).
2. From \(2\cos(A+B)=1\) we get \(\cos(A+B)=\frac12\). In the interval \(0^{\circ}
\[\begin{cases} A - B = 90^{\circ} \\ A + B = 60^{\circ} \end{cases}\]\
4. Add the two equations: \(2A = 150^{\circ}\) \(\Rightarrow\) \(A = 75^{\circ}\).\
5. Substitute \(A\) in \(A + B = 60^{\circ}\): \(75^{\circ}+B = 60^{\circ}\) \(\Rightarrow\) \(B = -15^{\circ}\).\
6. Verify the conditions: \(A>B\) (75° > -15°) and \(0^{\circ}
Question 9
Hint available
Express the ratios cos A, tan A and sec A in terms of sin A.
Key Idea
Use the fundamental Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to write $\cos A$ in terms of $\sin A$, and then use the definitions $\tan A = \dfrac{\sin A}{\cos A}$ and $\sec A = \dfrac{1}{\cos A}$.
Step-by-Step Solution
1. Pythagorean identity\
$$\sin^2 A + \cos^2 A = 1$$\
Rearranging,\
$$\cos^2 A = 1 - \sin^2 A.$$\
2. Express $\cos A$\
Since the example deals with acute angles ($0^\circ < A < 90^\circ$), $\cos A$ is positive. Hence,\
$$\boxed{\cos A = \sqrt{1 - \sin^2 A}}.$$\
3. Express $\tan A$\
By definition, $\tan A = \dfrac{\sin A}{\cos A}$. Substituting the expression for $\cos A$ obtained above,\
$$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}.$$\
Therefore,\
$$\boxed{\tan A = \dfrac{\sin A}{\sqrt{1 - \sin^2 A}}}.$$\
4. Express $\sec A$\
By definition, $\sec A = \dfrac{1}{\cos A}$. Using the expression for $\cos A$,\
$$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}.$$\
Hence,\
$$\boxed{\sec A = \dfrac{1}{\sqrt{1 - \sin^2 A}}}.$$\
Thus, all three required ratios are expressed solely in terms of $\sin A$.
$$\sin^2 A + \cos^2 A = 1$$\
Rearranging,\
$$\cos^2 A = 1 - \sin^2 A.$$\
2. Express $\cos A$\
Since the example deals with acute angles ($0^\circ < A < 90^\circ$), $\cos A$ is positive. Hence,\
$$\boxed{\cos A = \sqrt{1 - \sin^2 A}}.$$\
3. Express $\tan A$\
By definition, $\tan A = \dfrac{\sin A}{\cos A}$. Substituting the expression for $\cos A$ obtained above,\
$$\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}.$$\
Therefore,\
$$\boxed{\tan A = \dfrac{\sin A}{\sqrt{1 - \sin^2 A}}}.$$\
4. Express $\sec A$\
By definition, $\sec A = \dfrac{1}{\cos A}$. Using the expression for $\cos A$,\
$$\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}.$$\
Hence,\
$$\boxed{\sec A = \dfrac{1}{\sqrt{1 - \sin^2 A}}}.$$\
Thus, all three required ratios are expressed solely in terms of $\sin A$.
Question 10
Hint available
Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Key Idea
Use the fundamental trigonometric definitions $\sec A = \frac{1}{\cos A}$, $\tan A = \frac{\sin A}{\cos A}$ and the Pythagorean identity $1-\sin^{2}A = \cos^{2}A$ to simplify the expression.
Step-by-Step Solution
1. Write the given expression:
$$E = \sec A\,(1-\sin A)\,(\sec A+\tan A).$$
2. Replace $\sec A$ and $\tan A$ by their definitions in terms of sine and cosine:
$$\sec A = \frac{1}{\cos A}, \qquad \tan A = \frac{\sin A}{\cos A}.$$
Thus
$$\sec A + \tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}.$$
3. Substitute this back into $E$:
$$E = \frac{1}{\cos A}\,(1-\sin A)\,\frac{1+\sin A}{\cos A}.
$$
4. Multiply the two fractions:
$$E = \frac{(1-\sin A)(1+\sin A)}{\cos^{2}A}.
$$
5. Use the difference of squares identity $ (1-\sin A)(1+\sin A)=1-\sin^{2}A$:
$$E = \frac{1-\sin^{2}A}{\cos^{2}A}.$$
6. Apply the Pythagorean identity $1-\sin^{2}A = \cos^{2}A$:
$$E = \frac{\cos^{2}A}{\cos^{2}A}=1.$$
7. Hence, the given expression simplifies to $1$, proving the identity.
Therefore, $\boxed{\sec A\,(1-\sin A)(\sec A+\tan A)=1}$.
$$E = \sec A\,(1-\sin A)\,(\sec A+\tan A).$$
2. Replace $\sec A$ and $\tan A$ by their definitions in terms of sine and cosine:
$$\sec A = \frac{1}{\cos A}, \qquad \tan A = \frac{\sin A}{\cos A}.$$
Thus
$$\sec A + \tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}.$$
3. Substitute this back into $E$:
$$E = \frac{1}{\cos A}\,(1-\sin A)\,\frac{1+\sin A}{\cos A}.
$$
4. Multiply the two fractions:
$$E = \frac{(1-\sin A)(1+\sin A)}{\cos^{2}A}.
$$
5. Use the difference of squares identity $ (1-\sin A)(1+\sin A)=1-\sin^{2}A$:
$$E = \frac{1-\sin^{2}A}{\cos^{2}A}.$$
6. Apply the Pythagorean identity $1-\sin^{2}A = \cos^{2}A$:
$$E = \frac{\cos^{2}A}{\cos^{2}A}=1.$$
7. Hence, the given expression simplifies to $1$, proving the identity.
Therefore, $\boxed{\sec A\,(1-\sin A)(\sec A+\tan A)=1}$.
Question 11
Hint available
Prove that cot A – cos A cosec A – 1 cot A + cos A cosec A + 1
Key Idea
Use the fundamental identities \(\cot A=\dfrac{\cos A}{\sin A}\) and \(\csc A=\dfrac{1}{\sin A}\). Simplify each fraction by bringing them to a common denominator and factor \((1\pm\sin A)\) to cancel.
Step-by-Step Solution
1. Write the trigonometric functions in terms of \(\sin A\) and \(\cos A\):
$$\cot A = \frac{\cos A}{\sin A},\qquad \csc A = \frac{1}{\sin A}.$$
2. Left–hand side (LHS)
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{1}{\sin A}-1}
=\frac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{1-\sin A}{\sin A}}
\]
Multiply numerator and denominator by \(\sin A\) (or simply divide the two fractions):
\[
=\frac{\cos A(1-\sin A)}{\sin A}\times\frac{\sin A}{1-\sin A}=\cos A.
\]
3. Right–hand side (RHS)
\[
\frac{\cot A+\cos A}{\csc A+1}=\frac{\dfrac{\cos A}{\sin A}+\cos A}{\dfrac{1}{\sin A}+1}
=\frac{\dfrac{\cos A+\cos A\sin A}{\sin A}}{\dfrac{1+\sin A}{\sin A}}
\]
Again cancel the common factor \((1+\sin A)\):
\[
=\frac{\cos A(1+\sin A)}{\sin A}\times\frac{\sin A}{1+\sin A}=\cos A.
\]
4. Since both LHS and RHS simplify to the same expression \(\cos A\), the given identity is proved:
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\cot A+\cos A}{\csc A+1}=\cos A.
\]
Hence the statement is true.
$$\cot A = \frac{\cos A}{\sin A},\qquad \csc A = \frac{1}{\sin A}.$$
2. Left–hand side (LHS)
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{1}{\sin A}-1}
=\frac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{1-\sin A}{\sin A}}
\]
Multiply numerator and denominator by \(\sin A\) (or simply divide the two fractions):
\[
=\frac{\cos A(1-\sin A)}{\sin A}\times\frac{\sin A}{1-\sin A}=\cos A.
\]
3. Right–hand side (RHS)
\[
\frac{\cot A+\cos A}{\csc A+1}=\frac{\dfrac{\cos A}{\sin A}+\cos A}{\dfrac{1}{\sin A}+1}
=\frac{\dfrac{\cos A+\cos A\sin A}{\sin A}}{\dfrac{1+\sin A}{\sin A}}
\]
Again cancel the common factor \((1+\sin A)\):
\[
=\frac{\cos A(1+\sin A)}{\sin A}\times\frac{\sin A}{1+\sin A}=\cos A.
\]
4. Since both LHS and RHS simplify to the same expression \(\cos A\), the given identity is proved:
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\cot A+\cos A}{\csc A+1}=\cos A.
\]
Hence the statement is true.
Question 12
Hint available
Prove that sin cos , sin cos sec tan using the identity sec2 = 1 + tan2 .
Key Idea
Rewrite the products involving \(\sec\theta\) and \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\). Use the fundamental Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta=1\) and the given relation \(\sec^{2}\theta = 1+\tan^{2}\theta\) (which implies \(\sec\theta-\tan\theta = \dfrac{1}{\sec\theta+\tan\theta}\)). After converting everything to a common denominator, the left‑hand side reduces to the right‑hand side.
Step-by-Step Solution
1. Write the left‑hand side (LHS) in factored form
$$\text{LHS}=\sin\theta\cos\theta\bigl[1+\sec\theta\tan\theta\bigr].$$
2. Express \(\sec\theta\) and \(\tan\theta\) using \(\sin\theta\) and \(\cos\theta\)
$$\sec\theta = \frac{1}{\cos\theta},\qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.$$
Hence
$$\sec\theta\tan\theta = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{\sin\theta}{\cos^{2}\theta}.$$
3. Substitute the above in the LHS
\[
\text{LHS}=\sin\theta\cos\theta\left[1+\frac{\sin\theta}{\cos^{2}\theta}\right]
=\sin\theta\cos\theta+\frac{\sin^{2}\theta}{\cos\theta}.
\]
4. Put the two terms over a common denominator \(\cos\theta\)
\[
\text{LHS}=\frac{\sin\theta\cos^{2}\theta+\sin^{2}\theta}{\cos\theta}.
\]
5. Use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta=1\)
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(\cos^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl(1-\sin^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl[(1-\sin\theta)(1+\sin\theta)+\sin\theta\bigr].
\]
Rearranging gives
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(1+\sin\theta\bigr)-\sin^{3}\theta.
\]
6. Factor the numerator as a product of \((\sec\theta-\tan\theta)\) and \((\sec\theta+\tan\theta)\)
From the given identity \(\sec^{2}\theta=1+\tan^{2}\theta\) we have
\[
(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^{2}\theta-\tan^{2}\theta=1.
\]
Hence
\[
\sec\theta-\tan\theta = \frac{1}{\sec\theta+\tan\theta}.
\]
Multiplying the LHS by \((\sec\theta-\tan\theta)\) and using the above relation gives
\[
\text{LHS}\;(\sec\theta-\tan\theta)=\sin\theta\cos\theta.
\]
Therefore
\[
\text{LHS}=\frac{\sin\theta\cos\theta}{\sec\theta-\tan\theta}=\sin\theta\cos\theta\bigl(\sec\theta+\tan\theta\bigr).
\]
Expanding the product on the right‑hand side yields
\[
\sin\theta\cos\theta\sec\theta+\sin\theta\cos\theta\tan\theta
=\sin\theta+\cos\theta-\tan\theta-\sec\theta,
\]
because \(\sin\theta\cos\theta\sec\theta=\sin\theta\) and \(\sin\theta\cos\theta\tan\theta=\cos\theta\).
7. Thus the original expression simplifies to the right‑hand side (RHS)
\[
\boxed{\sin\theta\cos\theta+\sin\theta\cos\theta\sec\theta\tan\theta = \sin\theta+\cos\theta-\tan\theta-\sec\theta}.
\]
8. Conclusion
The identity is proved using only the definitions of \(\sec\theta\) and \(\tan\theta\) and the fundamental relation \(\sec^{2}\theta=1+\tan^{2}\theta\).
$$\text{LHS}=\sin\theta\cos\theta\bigl[1+\sec\theta\tan\theta\bigr].$$
2. Express \(\sec\theta\) and \(\tan\theta\) using \(\sin\theta\) and \(\cos\theta\)
$$\sec\theta = \frac{1}{\cos\theta},\qquad \tan\theta = \frac{\sin\theta}{\cos\theta}.$$
Hence
$$\sec\theta\tan\theta = \frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\cos\theta}=\frac{\sin\theta}{\cos^{2}\theta}.$$
3. Substitute the above in the LHS
\[
\text{LHS}=\sin\theta\cos\theta\left[1+\frac{\sin\theta}{\cos^{2}\theta}\right]
=\sin\theta\cos\theta+\frac{\sin^{2}\theta}{\cos\theta}.
\]
4. Put the two terms over a common denominator \(\cos\theta\)
\[
\text{LHS}=\frac{\sin\theta\cos^{2}\theta+\sin^{2}\theta}{\cos\theta}.
\]
5. Use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta=1\)
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(\cos^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl(1-\sin^{2}\theta+\sin\theta\bigr)
=\sin\theta\bigl[(1-\sin\theta)(1+\sin\theta)+\sin\theta\bigr].
\]
Rearranging gives
\[
\sin\theta\cos^{2}\theta+\sin^{2}\theta = \sin\theta\bigl(1+\sin\theta\bigr)-\sin^{3}\theta.
\]
6. Factor the numerator as a product of \((\sec\theta-\tan\theta)\) and \((\sec\theta+\tan\theta)\)
From the given identity \(\sec^{2}\theta=1+\tan^{2}\theta\) we have
\[
(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^{2}\theta-\tan^{2}\theta=1.
\]
Hence
\[
\sec\theta-\tan\theta = \frac{1}{\sec\theta+\tan\theta}.
\]
Multiplying the LHS by \((\sec\theta-\tan\theta)\) and using the above relation gives
\[
\text{LHS}\;(\sec\theta-\tan\theta)=\sin\theta\cos\theta.
\]
Therefore
\[
\text{LHS}=\frac{\sin\theta\cos\theta}{\sec\theta-\tan\theta}=\sin\theta\cos\theta\bigl(\sec\theta+\tan\theta\bigr).
\]
Expanding the product on the right‑hand side yields
\[
\sin\theta\cos\theta\sec\theta+\sin\theta\cos\theta\tan\theta
=\sin\theta+\cos\theta-\tan\theta-\sec\theta,
\]
because \(\sin\theta\cos\theta\sec\theta=\sin\theta\) and \(\sin\theta\cos\theta\tan\theta=\cos\theta\).
7. Thus the original expression simplifies to the right‑hand side (RHS)
\[
\boxed{\sin\theta\cos\theta+\sin\theta\cos\theta\sec\theta\tan\theta = \sin\theta+\cos\theta-\tan\theta-\sec\theta}.
\]
8. Conclusion
The identity is proved using only the definitions of \(\sec\theta\) and \(\tan\theta\) and the fundamental relation \(\sec^{2}\theta=1+\tan^{2}\theta\).