Exercise 9.1: Trigonometric Ratios
- Verify whether each of the following mathematical statements are true or false. Give argument in favour of your answer.
Solution:
a) The value of $\tan A$ is always less than $1$.
Answer: False.
Argument: In a right-angled triangle, the length of the perpendicular can be greater than the base. For example: $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is greater than $1$. Therefore, the value of $\tan A$ can also be greater than $1$.
b) $\cot A$ is the multiplication of $\cot$ and $A$.
Answer: False.
Argument: $\cot A$ denotes the cotangent ratio of angle $A$. It is not the product of $\cot$ and $A$; without $A$, $\cot$ alone is meaningless.
c) For any value of $A$, $\sec A = \frac{12}{5}$.
Answer: True.
Argument: We know that $\sec A = \frac{\text{Hypotenuse}}{\text{Base}}$. In a right-angled triangle, the hypotenuse is always greater than the base. Since $\frac{12}{5} > 1$, it is possible for $\sec A = \frac{12}{5}$ for an acute angle value of $A$.
d) $\cos$ is the smallest form of $\text{cotangent}$.
Answer: False.
Argument: $\cos$ is the abbreviated form of $\text{cosine}$. And the abbreviated form of $\text{cotangent}$ is $\cot$.
- If $\sin A = \frac{3}{4}$, find the other trigonometric ratios of angle $A$.
Solution:
Let $ABC$ be a right-angled triangle, where $\angle C = 90^\circ$ and $\angle A$ is an acute angle.
(Figure will be here)
Given that,
$\sin A = \frac{3}{4} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$
Therefore, Perpendicular $= 3$ and Hypotenuse $= 4$
According to Pythagoras’ theorem,
$\text{Base} = \sqrt{(\text{Hypotenuse})^2 – (\text{Perpendicular})^2}$
$= \sqrt{4^2 – 3^2}$
$= \sqrt{16 – 9}$
$= \sqrt{7}$
Other trigonometric ratios:
• $\cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{\sqrt{7}}{4}$
• $\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{\sqrt{7}}$
• $\cot A = \frac{\text{Base}}{\text{Perpendicular}} = \frac{\sqrt{7}}{3}$
• $\sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{4}{\sqrt{7}}$
• $\text{cosec } A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{4}{3}$
- Given that, $15\cot A = 8$, find the values of $\sin A$ and $\sec A$.
Solution:
Given that, $15\cot A = 8$
or, $\cot A = \frac{8}{15} = \frac{\text{Base}}{\text{Perpendicular}}$
Therefore, Base $= 8$ and Perpendicular $= 15$
According to Pythagoras’ theorem,
$\text{Hypotenuse} = \sqrt{(\text{Perpendicular})^2 + (\text{Base})^2}$
$= \sqrt{15^2 + 8^2}$
$= \sqrt{225 + 64}$
$= \sqrt{289} = 17$
Hence,
$\sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{15}{17}$
$\sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{17}{8}$
Answer: $\sin A = \frac{15}{17}$ and $\sec A = \frac{17}{8}$.
- If $\angle C$ is the right angle of the right-angled triangle $ABC$, $AB = 13\text{ cm}$, $BC = 12\text{ cm}$, and $\angle ABC = \theta$, find the values of $\sin\theta$, $\cos\theta$, and $\tan\theta$.
Solution:
(Figure will be here)
Given that, in the right-angled triangle $ABC$, $\angle C = 90^\circ$
Hypotenuse $AB = 13\text{ cm}$ and Base $BC = 12\text{ cm}$
With respect to angle $\angle ABC = \theta$, Perpendicular $= AC$
According to Pythagoras’ theorem,
$AC = \sqrt{AB^2 – BC^2}$
$= \sqrt{13^2 – 12^2}$
$= \sqrt{169 – 144}$
$= \sqrt{25}$
$= 5\text{ cm}$
Therefore, with respect to angle $\theta$:
• $\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{5}{13}$
• $\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{12}{13}$
• $\tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AC}{BC} = \frac{5}{12}$
- $\angle B$ is a right-angle of the right-angled triangle $ABC$, . If $\tan A = \sqrt{3}$, verify the truth of $\sqrt{3}\sin A\cos A = \frac{3}{4}$.
Solution:
Given that, in the right-angled triangle $ABC$, $\angle B = 90^\circ$
and $\tan A = \sqrt{3}$
We know that, $\tan 60^\circ = \sqrt{3}$
Therefore, $\tan A = \tan 60^\circ \implies A = 60^\circ$
Now, Left-Hand Side (L.H.S.) $= \sqrt{3}\sin A\cos A$
$= \sqrt{3}\sin 60^\circ\cos 60^\circ$
$= \sqrt{3} \times \frac{\sqrt{3}}{2} \times \frac{1}{2}$
$= \frac{3}{4} = \text{Right-Hand Side (R.H.S.)}$
Therefore, the statement $\sqrt{3}\sin A\cos A = \frac{3}{4}$ is true. (Verified)
Prove that (6-20):
6.
1) $\frac{1}{\sec^2 A} + \frac{1}{\text{cosec}^2 A} = 1$
Solution:
L.H.S. $= \frac{1}{\sec^2 A} + \frac{1}{\text{cosec}^2 A}$
$= \cos^2 A + \sin^2 A \quad \left[\because \frac{1}{\sec A} = \cos A, \frac{1}{\text{cosec } A} = \sin A\right]$
$= \sin^2 A + \cos^2 A$
$= 1 \quad \left[\because \sin^2 A + \cos^2 A = 1\right]$
$= \text{R.H.S.}$ (Proved)
2) $\frac{1}{\cos^2 A} – \frac{1}{\cot^2 A} = 1$
Solution:
L.H.S. $= \frac{1}{\cos^2 A} – \frac{1}{\cot^2 A}$
$= \sec^2 A – \tan^2 A \quad \left[\because \frac{1}{\cos A} = \sec A, \frac{1}{\cot A} = \tan A\right]$
$= 1 \quad \left[\because \sec^2 A – \tan^2 A = 1\right]$
$= \text{R.H.S.}$ (Proved)
3) $\frac{1}{\sin^2 A} – \frac{1}{\tan^2 A} = 1$
Solution:
L.H.S. $= \frac{1}{\sin^2 A} – \frac{1}{\tan^2 A}$
$= \text{cosec}^2 A – \cot^2 A \quad \left[\because \frac{1}{\sin A} = \text{cosec } A, \frac{1}{\tan A} = \cot A\right]$
$= 1 \quad \left[\because \text{cosec}^2 A – \cot^2 A = 1\right]$
$= \text{R.H.S.}$ (Proved)
7. 1) $\frac{\sin A}{\text{cosec } A} + \frac{\cos A}{\sec A} = 1$
Solution:
L.H.S. $= \frac{\sin A}{\text{cosec } A} + \frac{\cos A}{\sec A}$
$= \sin A \cdot \frac{1}{\text{cosec } A} + \cos A \cdot \frac{1}{\sec A}$
$= \sin A \cdot \sin A + \cos A \cdot \cos A \quad \left[\because \frac{1}{\text{cosec } A} = \sin A, \frac{1}{\sec A} = \cos A\right]$
$= \sin^2 A + \cos^2 A$
$= 1 \quad \left[\because \sin^2 A + \cos^2 A = 1\right]$
$= \text{R.H.S.}$ (Proved)
2) $\frac{\sec A}{\cos A} – \frac{\tan A}{\cot A} = 1$
Solution:
L.H.S. $= \frac{\sec A}{\cos A} – \frac{\tan A}{\cot A}$
$= \sec A \cdot \frac{1}{\cos A} – \tan A \cdot \frac{1}{\cot A}$
$= \sec A \cdot \sec A – \tan A \cdot \tan A \quad \left[\because \frac{1}{\cos A} = \sec A, \frac{1}{\cot A} = \tan A\right]$
$= \sec^2 A – \tan^2 A$
$= 1 \quad \left[\because \sec^2 A – \tan^2 A = 1\right]$
$= \text{R.H.S.}$ (Proved)
3) $\frac{1}{1 + \sin^2 A} + \frac{1}{1 + \text{cosec}^2 A} = 1$
Solution:
L.H.S. $= \frac{1}{1 + \sin^2 A} + \frac{1}{1 + \text{cosec}^2 A}$
$= \frac{1}{1 + \sin^2 A} + \frac{1}{1 + \frac{1}{\sin^2 A}}$
$= \frac{1}{1 + \sin^2 A} + \frac{1}{\frac{\sin^2 A + 1}{\sin^2 A}}$
$= \frac{1}{1 + \sin^2 A} + \frac{\sin^2 A}{1 + \sin^2 A}$
$= \frac{1 + \sin^2 A}{1 + \sin^2 A}$
$= 1$
$= \text{R.H.S.}$ (Proved)
8.
1) $\frac{\tan A}{1 – \cot A} + \frac{\cot A}{1 – \tan A} = \sec A \cdot \text{cosec } A + 1$
Solution:
L.H.S. $= \frac{\tan A}{1 – \cot A} + \frac{\cot A}{1 – \tan A}$
$= \frac{\frac{\sin A}{\cos A}}{1 – \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 – \frac{\sin A}{\cos A}}$
$= \frac{\frac{\sin A}{\cos A}}{\frac{\sin A – \cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{\frac{\cos A – \sin A}{\cos A}}$
$= \frac{\sin A}{\cos A} \cdot \frac{\sin A}{\sin A – \cos A} + \frac{\cos A}{\sin A} \cdot \frac{\cos A}{-(\sin A – \cos A)}$
$= \frac{\sin^2 A}{\cos A(\sin A – \cos A)} – \frac{\cos^2 A}{\sin A(\sin A – \cos A)}$
$= \frac{\sin^3 A – \cos^3 A}{\sin A \cos A(\sin A – \cos A)}$
$= \frac{(\sin A – \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A \cos A(\sin A – \cos A)}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A} \quad \left[\because \sin^2 A + \cos^2 A = 1\right]$
$= \frac{1}{\sin A \cos A} + \frac{\sin A \cos A}{\sin A \cos A}$
$= \frac{1}{\cos A} \cdot \frac{1}{\sin A} + 1$
$= \sec A \cdot \text{cosec } A + 1$$
$= \text{R.H.S.}$ (Proved)
2) $\frac{1}{1 + \tan^2 A} + \frac{1}{1 + \cot^2 A} = 1$
Solution:
L.H.S. $= \frac{1}{1 + \tan^2 A} + \frac{1}{1 + \cot^2 A}$
$= \frac{1}{\sec^2 A} + \frac{1}{\text{cosec}^2 A} \quad \left[\because 1 + \tan^2 A = \sec^2 A, 1 + \cot^2 A = \text{cosec}^2 A\right]$
$= \cos^2 A + \sin^2 A$
$= 1$
$= \text{R.H.S.}$ (Proved)
9. $\frac{\cos A}{1 – \tan A} + \frac{\sin A}{1 – \cot A} = \sin A + \cos A$
Solution:
L.H.S. $= \frac{\cos A}{1 – \tan A} + \frac{\sin A}{1 – \cot A}$
$= \frac{\cos A}{1 – \frac{\sin A}{\cos A}} + \frac{\sin A}{1 – \frac{\cos A}{\sin A}}$
$= \frac{\cos A}{\frac{\cos A – \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A – \cos A}{\sin A}}$
$= \frac{\cos^2 A}{\cos A – \sin A} + \frac{\sin^2 A}{\sin A – \cos A}$
$= \frac{\cos^2 A}{\cos A – \sin A} – \frac{\sin^2 A}{\cos A – \sin A}$
$= \frac{\cos^2 A – \sin^2 A}{\cos A – \sin A}$
$= \frac{(\cos A + \sin A)(\cos A – \sin A)}{\cos A – \sin A}$
$= \cos A + \sin A$
$= \sin A + \cos A$
$= \text{R.H.S.}$ (Proved)
10. $\tan A \sqrt{1 – \sin^2 A} = \sin A$
Solution:
L.H.S. $= \tan A \sqrt{1 – \sin^2 A}$
$= \tan A \cdot \sqrt{\cos^2 A} \quad \left[\because 1 – \sin^2 A = \cos^2 A\right]$
$= \tan A \cdot \cos A$
$= \frac{\sin A}{\cos A} \cdot \cos A$
$= \sin A$
$= \text{R.H.S.}$ (Proved)
11. $\frac{\sec A + \tan A}{\text{cosec } A + \cot A} = \frac{\text{cosec } A – \cot A}{\sec A – \tan A}$
Solution:
L.H.S. $= \frac{\sec A + \tan A}{\text{cosec } A + \cot A}$
$= \frac{(\sec A + \tan A)(\sec A – \tan A)(\text{cosec } A – \cot A)}{(\text{cosec } A + \cot A)(\text{cosec } A – \cot A)(\sec A – \tan A)}$ [Multiplying numerator and denominator by $(\sec A – \tan A)(\text{cosec } A – \cot A)$]
$= \frac{(\sec^2 A – \tan^2 A)(\text{cosec } A – \cot A)}{(\text{cosec}^2 A – \cot^2 A)(\sec A – \tan A)}$
$= \frac{1 \cdot (\text{cosec } A – \cot A)}{1 \cdot (\sec A – \tan A)} \quad \left[\because \sec^2 A – \tan^2 A = 1, \text{cosec}^2 A – \cot^2 A = 1\right]$
$= \frac{\text{cosec } A – \cot A}{\sec A – \tan A}$
$= \text{R.H.S.}$ (Proved)
12. $\frac{\text{cosec } A}{\text{cosec } A – 1} + \frac{\text{cosec } A}{\text{cosec } A + 1} = 2\sec^2 A$
Solution:
L.H.S. $= \frac{\text{cosec } A}{\text{cosec } A – 1} + \frac{\text{cosec } A}{\text{cosec } A + 1}$
$= \text{cosec } A \left[\frac{1}{\text{cosec } A – 1} + \frac{1}{\text{cosec } A + 1}\right]$
$= \text{cosec } A \left[\frac{(\text{cosec } A + 1) + (\text{cosec } A – 1)}{(\text{cosec } A – 1)(\text{cosec } A + 1)}\right]$
$= \text{cosec } A \cdot \frac{2\text{cosec } A}{\text{cosec}^2 A – 1}$
$= \frac{2\text{cosec}^2 A}{\cot^2 A} \quad \left[\because \text{cosec}^2 A – 1 = \cot^2 A\right]$
$= \frac{2 \cdot \frac{1}{\sin^2 A}}{\frac{\cos^2 A}{\sin^2 A}}$
$= \frac{2}{\sin^2 A} \cdot \frac{\sin^2 A}{\cos^2 A}$
$= \frac{2}{\cos^2 A}$
$= 2\sec^2 A$
$= \text{R.H.S.}$ (Proved)
13. $\frac{1}{1 + \sin A} + \frac{1}{1 – \sin A} = 2\sec^2 A$
Solution:
L.H.S. $= \frac{1}{1 + \sin A} + \frac{1}{1 – \sin A}$
$= \frac{(1 – \sin A) + (1 + \sin A)}{(1 + \sin A)(1 – \sin A)}$
$= \frac{2}{1 – \sin^2 A}$
$= \frac{2}{\cos^2 A} \quad \left[\because 1 – \sin^2 A = \cos^2 A\right]$
$= 2\sec^2 A$
$= \text{R.H.S.}$ (Proved)
14. $\frac{1}{\text{cosec } A – 1} – \frac{1}{\text{cosec } A + 1} = 2\tan^2 A$
Solution:
L.H.S. $= \frac{1}{\text{cosec } A – 1} – \frac{1}{\text{cosec } A + 1}$
$= \frac{(\text{cosec } A + 1) – (\text{cosec } A – 1)}{(\text{cosec } A – 1)(\text{cosec } A + 1)}$
$= \frac{\text{cosec } A + 1 – \text{cosec } A + 1}{\text{cosec}^2 A – 1}$
$= \frac{2}{\cot^2 A} \quad \left[\because \text{cosec}^2 A – 1 = \cot^2 A\right]$
$= 2\tan^2 A \quad \left[\because \frac{1}{\cot A} = \tan A\right]$
$= \text{R.H.S.}$ (Proved)
15. $\frac{\sin A}{1 – \cos A} + \frac{1 – \cos A}{\sin A} = 2\text{cosec } A$
Solution:
L.H.S. $= \frac{\sin A}{1 – \cos A} + \frac{1 – \cos A}{\sin A}$
$= \frac{\sin^2 A + (1 – \cos A)^2}{\sin A(1 – \cos A)}$
$= \frac{\sin^2 A + 1 – 2\cos A + \cos^2 A}{\sin A(1 – \cos A)}$
$= \frac{(\sin^2 A + \cos^2 A) + 1 – 2\cos A}{\sin A(1 – \cos A)}$
$= \frac{1 + 1 – 2\cos A}{\sin A(1 – \cos A)} \quad \left[\because \sin^2 A + \cos^2 A = 1\right]$
$= \frac{2 – 2\cos A}{\sin A(1 – \cos A)}$
$= \frac{2(1 – \cos A)}{\sin A(1 – \cos A)}$
$= \frac{2}{\sin A}$
$= 2\text{cosec } A$
$= \text{R.H.S.}$ (Proved)
16. $\frac{\tan A}{\sec A + 1} – \frac{\sec A – 1}{\tan A} = 0$
Solution:
L.H.S. $= \frac{\tan A}{\sec A + 1} – \frac{\sec A – 1}{\tan A}$
$= \frac{\tan^2 A – (\sec A – 1)(\sec A + 1)}{\tan A(\sec A + 1)}$
$= \frac{\tan^2 A – (\sec^2 A – 1)}{\tan A(\sec A + 1)}$
$= \frac{\tan^2 A – \tan^2 A}{\tan A(\sec A + 1)} \quad \left[\because \sec^2 A – 1 = \tan^2 A\right]$
$= \frac{0}{\tan A(\sec A + 1)}$
$= 0$
$= \text{R.H.S.}$ (Proved)
17. $(\tan\theta + \sec\theta)^2 = \frac{1 + \sin\theta}{1 – \sin\theta}$
Solution:
L.H.S. $= (\tan\theta + \sec\theta)^2$
$= \left(\frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta}\right)^2$
$= \left(\frac{\sin\theta + 1}{\cos\theta}\right)^2$
$= \frac{(1 + \sin\theta)^2}{\cos^2\theta}$
$= \frac{(1 + \sin\theta)^2}{1 – \sin^2\theta} \quad \left[\because \cos^2\theta = 1 – \sin^2\theta\right]$
$= \frac{(1 + \sin\theta)(1 + \sin\theta)}{(1 – \sin\theta)(1 + \sin\theta)}$
$= \frac{1 + \sin\theta}{1 – \sin\theta}$
$= \text{R.H.S.}$ (Proved)
18. $\frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \cdot \tan B$
Solution:
L.H.S. $= \frac{\cot A + \tan B}{\cot B + \tan A}$
$= \frac{\cot A + \tan B}{\frac{1}{\tan B} + \frac{1}{\cot A}}$
$= \frac{\cot A + \tan B}{\frac{\cot A + \tan B}{\tan B \cdot \cot A}}$
$= (\cot A + \tan B) \times \frac{\tan B \cdot \cot A}{\cot A + \tan B}$
$= \cot A \cdot \tan B$$
$= \text{R.H.S.}$ (Proved)
19. $\sqrt{\frac{1 – \sin A}{1 + \sin A}} = \sec A – \tan A$
Solution:
L.H.S. $= \sqrt{\frac{1 – \sin A}{1 + \sin A}}$
$= \frac{\sqrt{(1 – \sin A)(1 – \sin A)}}{\sqrt{(1 + \sin A)(1 – \sin A)}}$ [Multiplying numerator and denominator by $\sqrt{1 – \sin A}$]
$= \frac{\sqrt{(1 – \sin A)^2}}{\sqrt{1 – \sin^2 A}}$
$= \frac{1 – \sin A}{\sqrt{\cos^2 A}} \quad \left[\because 1 – \sin^2 A = \cos^2 A\right]$
$= \frac{1 – \sin A}{\cos A}$$
$= \frac{1}{\cos A} – \frac{\sin A}{\cos A}$
$= \sec A – \tan A$$
$= \text{R.H.S.}$ (Proved)
20. $\sqrt{\frac{\sec A + 1}{\sec A – 1}} = \cot A + \text{cosec } A$
Solution:
L.H.S. $= \sqrt{\frac{\sec A + 1}{\sec A – 1}}$
$= \frac{\sqrt{(\sec A + 1)(\sec A + 1)}}{\sqrt{(\sec A – 1)(\sec A + 1)}}$ [Multiplying numerator and denominator by $\sqrt{\sec A + 1}$]
$= \frac{\sqrt{(\sec A + 1)^2}}{\sqrt{\sec^2 A – 1}}$
$= \frac{\sec A + 1}{\sqrt{\tan^2 A}} \quad \left[\because \sec^2 A – 1 = \tan^2 A\right]$
$= \frac{\sec A + 1}{\tan A}$
$= \frac{\sec A}{\tan A} + \frac{1}{\tan A}$
$= \frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}} + \cot A$
$= \left(\frac{1}{\cos A} \times \frac{\cos A}{\sin A}\right) + \cot A$
$= \frac{1}{\sin A} + \cot A$
$= \text{cosec } A + \cot A$
$= \cot A + \text{cosec } A$
$= \text{R.H.S.}$ (Proved)
21. If $\cos A + \sin A = \sqrt{2}\cos A$, prove that, $\cos A – \sin A = \sqrt{2}\sin A$
Solution:
Given that, $\cos A + \sin A = \sqrt{2}\cos A$
or, $\sqrt{2}(\cos A + \sin A) = \sqrt{2} \cdot \sqrt{2}\cos A$ [Multiplying both sides by $\sqrt{2}$]
or, $\sqrt{2}\cos A + \sqrt{2}\sin A = 2\cos A$
[Substituting $(\cos A + \sin A)$ in place of $\sqrt{2}\cos A$]
or, $(\cos A + \sin A) + \sqrt{2}\sin A = 2\cos A$
or, $\sqrt{2}\sin A = 2\cos A – \cos A – \sin A$
or, $\sqrt{2}\sin A = \cos A – \sin A$
or, $\cos A – \sin A = \sqrt{2}\sin A$ (Proved)
22. If $\tan A = \frac{1}{\sqrt{3}}$, find the value of $\frac{\text{cosec}^2 A – \sec^2 A}{\text{cosec}^2 A + \sec^2 A}$.
Solution:
Given that, $\tan A = \frac{1}{\sqrt{3}}$
We know that, $\cot A = \frac{1}{\tan A} = \sqrt{3}$
Now,
$\sec^2 A = 1 + \tan^2 A$
$= 1 + \left(\frac{1}{\sqrt{3}}\right)^2$
$= 1 + \frac{1}{3}$
$= \frac{4}{3}$
and $\text{cosec}^2 A = 1 + \cot^2 A$
$= 1 + (\sqrt{3})^2$
$= 1 + 3 = 4$
Given expression $= \frac{\text{cosec}^2 A – \sec^2 A}{\text{cosec}^2 A + \sec^2 A}$
$= \frac{4 – \frac{4}{3}}{4 + \frac{4}{3}}$
$= \frac{\frac{12 – 4}{3}}{\frac{12 + 4}{3}}$
$= \frac{\frac{8}{3}}{\frac{16}{3}}$
$= \frac{8}{3} \times \frac{3}{16}$
$= \frac{1}{2}$
Answer: $\frac{1}{2}$
23. If $\text{cosec } A – \cot A = \frac{4}{3}$, what is the value of $\text{cosec } A + \cot A$?
Solution:
Given that, $\text{cosec } A – \cot A = \frac{4}{3}$
We know that, $\text{cosec}^2 A – \cot^2 A = 1$
or, $(\text{cosec } A + \cot A)(\text{cosec } A – \cot A) = 1$
$(\text{cosec } A + \cot A) \times \frac{4}{3} = 1$ [Substituting the value]
or, $\text{cosec } A + \cot A = \frac{1}{\frac{4}{3}}$
or, $\text{cosec } A + \cot A = \frac{3}{4}$
Answer: $\frac{3}{4}$
24. If $\cot A = \frac{b}{a}$, determine the value of $\frac{a\sin A – b\cos A}{a\sin A + b\cos A}$.
Solution:
Given that,
$\cot A = \frac{b}{a}$
or, $\frac{\cos A}{\sin A} = \frac{b}{a}$
$\frac{\sin A}{\cos A} = \frac{a}{b}$ [By invertendo]
$\frac{a\sin A}{b\cos A} = \frac{a \times a}{b \times b}$ [Multiplying both sides by $\frac{a}{b}$]
$\frac{a\sin A}{b\cos A} = \frac{a^2}{b^2}$
$\frac{a\sin A – b\cos A}{a\sin A + b\cos A} = \frac{a^2 – b^2}{a^2 + b^2}$ [By dividendo and componendo]
Answer: $\frac{a^2 – b^2}{a^2 + b^2}$
25. $\text{cosec} A – \cot A = x$ and $\text{cosec} A + \cot A = y$
a) If $\sin\theta = \frac{5}{13}$, find the value of $\sec\theta$.
Solution:
Given that,
$\sin\theta = \frac{5}{13}$
We know that,
$\cos\theta = \sqrt{1 – \sin^2\theta}$
$= \sqrt{1 – \left(\frac{5}{13}\right)^2}$
$= \sqrt{1 – \frac{25}{169}}$
$= \sqrt{\frac{144}{169}}$
$= \frac{12}{13}$
Therefore, $\sec\theta = \frac{1}{\cos\theta}$
$= \frac{1}{\frac{12}{13}}$
$= \frac{13}{12}$
Answer: $\frac{13}{12}$
b) Show that, $\sec A = \frac{1 + x^2}{1 – x^2}$
Solution:
Given that,
$\text{cosec} A – \cot A = x$
or, $\frac{1}{\sin A} – \frac{\cos A}{\sin A} = x$
or, $\frac{1 – \cos A}{\sin A} = x$
or, $\frac{(1 – \cos A)^2}{\sin^2 A} = x^2$ [Squaring both sides]
or, $\frac{(1 – \cos A)^2}{1 – \cos^2 A} = x^2 \quad \left[\because \sin^2 A = 1 – \cos^2 A\right]$
or, $\frac{(1 – \cos A)(1 – \cos A)}{(1 – \cos A)(1 + \cos A)} = x^2$
or, $\frac{1 – \cos A}{1 + \cos A} = x^2$
or, $\frac{(1 – \cos A) – (1 + \cos A)}{(1 – \cos A) + (1 + \cos A)} = \frac{x^2 – 1}{x^2 + 1}$ [By dividendo and componendo]
or, $\frac{1 – \cos A – 1 – \cos A}{1 – \cos A + 1 + \cos A} = \frac{x^2 – 1}{x^2 + 1}$
or, $\frac{-2\cos A}{2} = \frac{x^2 – 1}{x^2 + 1}$
or, $-\cos A = \frac{x^2 – 1}{x^2 + 1}$
or, $\cos A = -\left(\frac{x^2 – 1}{x^2 + 1}\right) = \frac{1 – x^2}{1 + x^2}$
Therefore,
$\sec A = \frac{1}{\cos A} = \frac{1 + x^2}{1 – x^2}$
(Shown)
c) If $\frac{x}{y} = 7 – 4\sqrt{3}$, find the value of $A$.
Solution:
Given that, $\text{cosec} A – \cot A = x$ and $\text{cosec} A + \cot A = y$
According to the question,
$\frac{x}{y} = 7 – 4\sqrt{3}$
or, $\frac{\text{cosec} A – \cot A}{\text{cosec} A + \cot A} = \frac{7 – 4\sqrt{3}}{1}$
$\frac{(\text{cosec} A – \cot A) – (\text{cosec} A + \cot A)}{(\text{cosec} A – \cot A) + (\text{cosec} A + \cot A)} = \frac{(7 – 4\sqrt{3}) – 1}{(7 – 4\sqrt{3}) + 1}$ [By dividendo and componendo]
or, $\frac{-2\cot A}{2\text{cosec} A} = \frac{6 – 4\sqrt{3}}{8 – 4\sqrt{3}}$
or, $\frac{\cot A}{\text{cosec} A} = \frac{2(3 – 2\sqrt{3})}{2(4 – 2\sqrt{3})}$
or, $\frac{\frac{\cos A}{\sin A}}{\frac{1}{\sin A}} = \frac{3 – 2\sqrt{3}}{4 – 2\sqrt{3}}$
or, $\cos A = \frac{\sqrt{3}(\sqrt{3} – 2)}{-2\sqrt{3}(\sqrt{3} – 2)}$
or, $\cos A = \frac{\sqrt{3}}{2\sqrt{3}}$
or, $\cos A = \frac{1}{2}$
We know that, $\cos 60^\circ = \frac{1}{2}$
Therefore, $\cos A = \cos 60^\circ \implies A = 60^\circ$
Answer: $A = 60^\circ$