Exercise 9.2: Trigonometric Ratios of Specific Angles
- If $\cos\theta = \frac{1}{2}$, what is the value of $\cot\theta$?
Solution:
Given, $\cos\theta = \frac{1}{2}$
or, $\cos\theta = \cos 60^\circ$
or, $\theta = 60^\circ$
Now, $\cot\theta = \cot 60^\circ = \frac{1}{\sqrt{3}}$
Answer: a) $\frac{1}{\sqrt{3}}$
- If $\cos^2\theta – \sin^2\theta = \frac{1}{3}$, what is the value of $\cos^4\theta – \sin^4\theta$?
Solution:
Given Expression $= \cos^4\theta – \sin^4\theta$
$= (\cos^2\theta)^2 – (\sin^2\theta)^2$
$= (\cos^2\theta + \sin^2\theta)(\cos^2\theta – \sin^2\theta)$
$= 1 \times \frac{1}{3} \quad \left[\because \cos^2\theta + \sin^2\theta = 1 \text{ and } \cos^2\theta – \sin^2\theta = \frac{1}{3}\right]$
$= \frac{1}{3}$
Answer: d) $\frac{1}{3}$
- If $\cot(\theta – 30^\circ) = \frac{1}{\sqrt{3}}$, what is the value of $\sin\theta$?
Solution:
Given, $\cot(\theta – 30^\circ) = \frac{1}{\sqrt{3}}$
or, $\cot(\theta – 30^\circ) = \cot 60^\circ$
or, $\theta – 30^\circ = 60^\circ$
or, $\theta = 60^\circ + 30^\circ$
or, $\theta = 90^\circ$
Now, $\sin\theta = \sin 90^\circ = 1$
Answer: c) $1$
- If $\tan(3A) = \sqrt{3}$, $A=$ what ?
Solution:
Given, $\tan(3A) = \sqrt{3}$
or, $\tan(3A) = \tan 60^\circ$
or, $3A = 60^\circ$
or, $A = \frac{60^\circ}{3}$
or, $A = 20^\circ$
Answer: c) $20^\circ$
Determine the value of (5 – 8):
- $\frac{1 – \cot^2 60^\circ}{1 + \cot^2 60^\circ}$
Solution:
We know, $\cot 60^\circ = \frac{1}{\sqrt{3}}$
Given Expression $= \frac{1 – \left(\frac{1}{\sqrt{3}}\right)^2}{1 + \left(\frac{1}{\sqrt{3}}\right)^2}$
$= \frac{1 – \frac{1}{3}}{1 + \frac{1}{3}}$
$= \frac{\frac{3 – 1}{3}}{\frac{3 + 1}{3}}$
$= \frac{\frac{2}{3}}{\frac{4}{3}}$
$= \frac{2}{3} \times \frac{3}{4}$
$= \frac{1}{2}$
Answer: $\frac{1}{2}$
- $\tan 45^\circ \cdot \sin^2 60^\circ \cdot \tan 30^\circ \cdot \tan 60^\circ$
Solution:
Given Expression $= \tan 45^\circ \cdot \sin^2 60^\circ \cdot \tan 30^\circ \cdot \tan 60^\circ$
$= 1 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \cdot \frac{1}{\sqrt{3}} \cdot \sqrt{3}$
$= 1 \cdot \frac{3}{4} \cdot 1$
$= \frac{3}{4}$
Answer: $\frac{3}{4}$
- $\frac{1 – \cos^2 60^\circ}{1 + \cos^2 60^\circ} + \sec^2 60^\circ$
Solution:
Given Expression $= \frac{1 – \cos^2 60^\circ}{1 + \cos^2 60^\circ} + \sec^2 60^\circ$
$= \frac{1 – \left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{2}\right)^2} + (2)^2$
$= \frac{1 – \frac{1}{4}}{1 + \frac{1}{4}} + 4$
$= \frac{\frac{3}{4}}{\frac{5}{4}} + 4$
$= \left(\frac{3}{4} \times \frac{4}{5}\right) + 4$
$= \frac{3}{5} + 4$
$= \frac{3 + 20}{5}$
$= \frac{23}{5}$
Answer: $\frac{23}{5}$
- $\cos 45^\circ \cdot \cot^2 60^\circ \cdot \text{cosec}^2 30^\circ$
Solution:
We know, $\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\cot 60^\circ = \frac{1}{\sqrt{3}}$, and $\text{cosec } 30^\circ = 2$
Given Expression $= \cos 45^\circ \cdot \cot^2 60^\circ \cdot \text{cosec}^2 30^\circ$
$= \frac{1}{\sqrt{2}} \cdot \left(\frac{1}{\sqrt{3}}\right)^2 \cdot (2)^2$
$= \frac{1}{\sqrt{2}} \cdot \frac{1}{3} \cdot 4$
$= \frac{4}{3\sqrt{2}}$
$= \frac{2 \times 2}{3\sqrt{2}}$
$= \frac{2\sqrt{2} \times \sqrt{2}}{3\sqrt{2}}$
$= \frac{2\sqrt{2}}{3}$
Answer: $\frac{2\sqrt{2}}{3}$
Show that (9 – 14):
- $\cos^2 30^\circ – \sin^2 30^\circ = \cos 60^\circ$
Solution:
L.H.S. $= \cos^2 30^\circ – \sin^2 30^\circ$
$= \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{1}{2}\right)^2$
$= \frac{3}{4} – \frac{1}{4}$
$= \frac{3 – 1}{4}$
$= \frac{2}{4}$
$= \frac{1}{2}$
R.H.S. $= \cos 60^\circ = \frac{1}{2}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- $\sin 60^\circ \cdot \cos 30^\circ + \cos 60^\circ \cdot \sin 30^\circ = \sin 90^\circ$
Solution:
L.H.S. $= \sin 60^\circ \cdot \cos 30^\circ + \cos 60^\circ \cdot \sin 30^\circ$
$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2}$
$= \frac{3}{4} + \frac{1}{4}$
$= \frac{3 + 1}{4} = \frac{4}{4} = 1$
R.H.S. $= \sin 90^\circ = 1$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- $\cos 60^\circ \cdot \cos 30^\circ + \sin 60^\circ \cdot \sin 30^\circ = \cos 30^\circ$
Solution:
L.H.S. $= \cos 60^\circ \cdot \cos 30^\circ + \sin 60^\circ \cdot \sin 30^\circ$
$= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2}$
$= \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}$
$= \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
R.H.S. $= \cos 30^\circ = \frac{\sqrt{3}}{2}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- $\sin 3A = \cos 3A$ when $A = 15^\circ$.
Solution:
Given, $A = 15^\circ$
L.H.S. $= \sin 3A$
$= \sin(3 \times 15^\circ)$
$= \sin 45^\circ$
$= \frac{1}{\sqrt{2}}$
R.H.S. $= \cos 3A$
$= \cos(3 \times 15^\circ)$
$= \cos 45^\circ$
$= \frac{1}{\sqrt{2}}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- $\sin 2A = \frac{2\tan A}{1 + \tan^2 A}$ when $A = 45^\circ$.
Solution:
Given, $A = 45^\circ$
L.H.S. $= \sin 2A$
$= \sin(2 \times 45^\circ)$
$= \sin 90^\circ = 1$
R.H.S. $= \frac{2\tan 45^\circ}{1 + \tan^2 45^\circ}$
$= \frac{2(1)}{1 + (1)^2}$
$= \frac{2}{1 + 1}$
$= \frac{2}{2} = 1$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- $\tan 2A = \frac{2\tan A}{1 – \tan^2 A}$ when $A = 30^\circ$.
Solution:
Given, $A = 30^\circ$
L.H.S. $= \tan 2A$
$= \tan(2 \times 30^\circ)$
$= \tan 60^\circ$
$= \sqrt{3}$
R.H.S. $= \frac{2\tan 30^\circ}{1 – \tan^2 30^\circ}$
$= \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 – \left(\frac{1}{\sqrt{3}}\right)^2}$
$= \frac{\frac{2}{\sqrt{3}}}{1 – \frac{1}{3}}$
$= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
$= \frac{2}{\sqrt{3}} \times \frac{3}{2}$
$= \sqrt{3}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- If $2\cos(A + B) = 1 = 2\sin(A – B)$ and $A, B$ are acute angles, show that $A = 45^\circ, B = 15^\circ$.
Solution:
Given,
$2\cos(A + B) = 1$
or, $\cos(A + B) = \frac{1}{2}$
or, $\cos(A + B) = \cos 60^\circ$
or, $A + B = 60^\circ$ …….. (1)
Again,
$2\sin(A – B) = 1$
or, $\sin(A – B) = \frac{1}{2}$
or, $\sin(A – B) = \sin 30^\circ$
or, $A – B = 30^\circ$ …….. (2)
Adding equations (1) and (2):
$2A = 90^\circ \implies A = 45^\circ$
Subtracting equation (2) from equation (1):
$2B = 30^\circ \implies B = 15^\circ$
Therefore, $A = 45^\circ$ and $B = 15^\circ$. (Shown)
- If $\cos(A – B) = 1$, $2\sin(A + B) = \sqrt{3}$ and $A, B$ are acute angles, determine the values of $A$ and $B$.
Solution:
Given,
$\cos(A – B) = 1$
or, $\cos(A – B) = \cos 0^\circ$
or, $A – B = 0^\circ$ …….. (1)
Again,
$2\sin(A + B) = \sqrt{3}$
or, $\sin(A + B) = \frac{\sqrt{3}}{2}$
or, $\sin(A + B) = \sin 60^\circ$
or, $A + B = 60^\circ$ …….. (2)
Adding equations (1) and (2):
$2A = 60^\circ \implies A = 30^\circ$
Subtracting equation (1) from equation (2):
$2B = 60^\circ \implies B = 30^\circ$
Answer: $A = 30^\circ, B = 30^\circ$
- Solve: $\frac{\cos A – \sin A}{\cos A + \sin A} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}$
Solution:
Given equation,
$\frac{\cos A – \sin A}{\cos A + \sin A} = \frac{\sqrt{3} – 1}{\sqrt{3} + 1}$
[By componendo and dividendo]
or, $\frac{(\cos A – \sin A) + (\cos A + \sin A)}{(\cos A – \sin A) – (\cos A + \sin A)} = \frac{(\sqrt{3} – 1) + (\sqrt{3} + 1)}{(\sqrt{3} – 1) – (\sqrt{3} + 1)}$
or, $\frac{2\cos A}{-2\sin A} = \frac{2\sqrt{3}}{-2}$
or, $\frac{\cos A}{\sin A} = \sqrt{3}$
or, $\cot A = \sqrt{3}$
or, $\cot A = \cot 30^\circ$
or, $A = 30^\circ$
Answer: $A = 30^\circ$
- If $A$ and $B$ are acute angles and $\cot(A + B) = 1$ and $\cot(A – B) = \sqrt{3}$, determine the values of $A$ and $B$.
Solution:
Given,
$\cot(A + B) = 1$
or, $\cot(A + B) = \cot 45^\circ$
or, $A + B = 45^\circ$ …….. (1)
and $\cot(A – B) = \sqrt{3}$
or, $\cot(A – B) = \cot 30^\circ$
or, $A – B = 30^\circ$ …….. (2)
Adding (1) and (2):
$2A = 75^\circ \implies A = 37.5^\circ \text{ or } 37\frac{1}{2}^\circ$
Subtracting (2) from (1):
$2B = 15^\circ \implies B = 7.5^\circ \text{ or } 7\frac{1}{2}^\circ$
Answer: $A = 37\frac{1}{2}^\circ, B = 7\frac{1}{2}^\circ$
- Show that $\cos 3A = 4\cos^3 A – 3\cos A$ when $A = 30^\circ$.
Solution:
Given, $A = 30^\circ$
L.H.S. $= \cos 3A$
$= \cos(3 \times 30^\circ)$
$= \cos 90^\circ = 0$
R.H.S. $= 4\cos^3 30^\circ – 3\cos 30^\circ$
$= 4 \left(\frac{\sqrt{3}}{2}\right)^3 – 3 \left(\frac{\sqrt{3}}{2}\right)$
$= 4 \left(\frac{3\sqrt{3}}{8}\right) – \frac{3\sqrt{3}}{2}$
$= \frac{3\sqrt{3}}{2} – \frac{3\sqrt{3}}{2} = 0$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
- Solve: $\sin\theta + \cos\theta = 1$, when $0^\circ \le \theta \le 90^\circ$
Solution:
$\sin\theta + \cos\theta = 1$
or, $\sin\theta = 1 – \cos\theta$
[Squaring both sides]
$\sin^2\theta = (1 – \cos\theta)^2$
or, $1 – \cos^2\theta = 1 – 2\cos\theta + \cos^2\theta$
or, $2\cos^2\theta – 2\cos\theta = 0$
or, $2\cos\theta(\cos\theta – 1) = 0$
Either,
$2\cos\theta = 0 \implies \cos\theta = 0 \implies \cos\theta = \cos 90^\circ \implies \theta = 90^\circ$
Or,
$\cos\theta – 1 = 0 \implies \cos\theta = 1 \implies \cos\theta = \cos 0^\circ \implies \theta = 0^\circ$
Answer: $\theta = 0^\circ \text{ or } 90^\circ$
- Solve: $\cos^2\theta – \sin^2\theta = 2 – 5\cos\theta$, when $\theta$ is an acute angle.
Solution:
$\cos^2\theta – \sin^2\theta = 2 – 5\cos\theta$
or, $\cos^2\theta – (1 – \cos^2\theta) = 2 – 5\cos\theta$
or, $\cos^2\theta – 1 + \cos^2\theta – 2 + 5\cos\theta = 0$
or, $2\cos^2\theta + 5\cos\theta – 3 = 0$
or, $2\cos^2\theta + 6\cos\theta – \cos\theta – 3 = 0$
or, $2\cos\theta(\cos\theta + 3) – 1(\cos\theta + 3) = 0$
or, $(\cos\theta + 3)(2\cos\theta – 1) = 0$
Since $\cos\theta + 3 \neq 0$ (because the value of $\cos\theta$ cannot lie outside the range $[-1, 1]$),
Therefore,
$2\cos\theta – 1 = 0$
or, $2\cos\theta = 1 \implies \cos\theta = \frac{1}{2} \implies \cos\theta = \cos 60^\circ \implies \theta = 60^\circ$
Answer: $\theta = 60^\circ$
- Solve: $2\sin^2\theta + 3\cos\theta – 3 = 0$, $\theta$ is an acute angle.
Solution:
$2\sin^2\theta + 3\cos\theta – 3 = 0$
or, $2(1 – \cos^2\theta) + 3\cos\theta – 3 = 0$
or, $2 – 2\cos^2\theta + 3\cos\theta – 3 = 0$
or, $-2\cos^2\theta + 3\cos\theta – 1 = 0$
or, $2\cos^2\theta – 3\cos\theta + 1 = 0$
or, $2\cos^2\theta – 2\cos\theta – \cos\theta + 1 = 0$
or, $2\cos\theta(\cos\theta – 1) – 1(\cos\theta – 1) = 0$
or, $(\cos\theta – 1)(2\cos\theta – 1) = 0$
Since $\theta$ is an acute angle, $\theta \neq 0^\circ$ (meaning $\cos\theta \neq 1$).
Therefore,
$2\cos\theta – 1 = 0 \implies \cos\theta = \frac{1}{2} \implies \theta = 60^\circ$
Answer: $\theta = 60^\circ$
- Solve: $\tan^2\theta – (1 + \sqrt{3})\tan\theta + \sqrt{3} = 0$
Solution:
$\tan^2\theta – (1 + \sqrt{3})\tan\theta + \sqrt{3} = 0$
or, $\tan^2\theta – \tan\theta – \sqrt{3}\tan\theta + \sqrt{3} = 0$
or, $\tan\theta(\tan\theta – 1) – \sqrt{3}(\tan\theta – 1) = 0$
or, $(\tan\theta – 1)(\tan\theta – \sqrt{3}) = 0$
Either,
$\tan\theta – 1 = 0 \implies \tan\theta = 1 \implies \tan\theta = \tan 45^\circ \implies \theta = 45^\circ$
Or,
$\tan\theta – \sqrt{3} = 0 \implies \tan\theta = \sqrt{3} \implies \tan\theta = \tan 60^\circ \implies \theta = 60^\circ$
Answer: $\theta = 45^\circ \text{ or } 60^\circ$
- Solve: $3\cot^2 60^\circ + \frac{1}{4}\text{cosec}^2 30^\circ + 5\sin^2 45^\circ – 4\cos^2 60^\circ$
Solution:
Given Expression $= 3\cot^2 60^\circ + \frac{1}{4}\text{cosec}^2 30^\circ + 5\sin^2 45^\circ – 4\cos^2 60^\circ$
$= 3\left(\frac{1}{\sqrt{3}}\right)^2 + \frac{1}{4}(2)^2 + 5\left(\frac{1}{\sqrt{2}}\right)^2 – 4\left(\frac{1}{2}\right)^2$
$= 3 \cdot \frac{1}{3} + \frac{1}{4} \cdot 4 + 5 \cdot \frac{1}{2} – 4 \cdot \frac{1}{4}$
$= 1 + 1 + \frac{5}{2} – 1$
$= 1 + \frac{5}{2}$
$= \frac{2 + 5}{2} = \frac{7}{2}$
Answer: $\frac{7}{2} \text{ or } 3\frac{1}{2}$
- If $\angle B = 90^\circ$, $AB = 5\text{ cm}$, $BC = 12\text{ cm}$ of $\triangle ABC$.
(a) Find the length of $AC$.
Solution:
According to Pythagoras’ theorem,
$AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$
Answer: $AC = 13\text{ cm}$
(b) If $\angle C = \theta$, find the value of $\sin\theta + \cos\theta$.
Solution:
With respect to angle $\angle C = \theta$:
Perpendicular $= AB = 5$, Base $= BC = 12$, Hypotenuse $= AC = 13$
Therefore,
$\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13}$
$\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{12}{13}$
Given Expression $= \sin\theta + \cos\theta$
$= \frac{5}{13} + \frac{12}{13}$
$= \frac{5 + 12}{13}$
$= \frac{17}{13}$
Answer: $\frac{17}{13}$
(c) Using stem show that $\sec^2 A + \text{cosec}^2 A = \sec^2 A \cdot \text{cosec}^2 A$
Solution:
With respect to angle $\angle A$:
Perpendicular $= BC = 12$, Base $= AB = 5$, Hypotenuse $= AC = 13$
Therefore,
$\sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{13}{5}$
$\text{cosec } A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{13}{12}$
L.H.S. $= \sec^2 A + \text{cosec}^2 A$
$= \left(\frac{13}{5}\right)^2 + \left(\frac{13}{12}\right)^2$
$= \frac{169}{25} + \frac{169}{144}$
$= \frac{169 \times 144 + 169 \times 25}{3600}$
$= \frac{169(144 + 25)}{3600}$
$= \frac{169 \times 169}{3600}$
$= \frac{28561}{3600}$
R.H.S. $= \sec^2 A \cdot \text{cosec}^2 A$
$= \left(\frac{13}{5}\right)^2 \cdot \left(\frac{13}{12}\right)^2$
$= \frac{169}{25} \cdot \frac{169}{144}$
$= \frac{28561}{3600}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)
Sample Questions: Multiple Choice Questions (MCQ)
- For $0^\circ \le \theta \le 90^\circ$, what is the maximum value of $\sin\theta$?
Solution:
$\sin 0^\circ = 0$, $\sin 30^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{1}{\sqrt{2}}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\sin 90^\circ = 1$
So, within $0^\circ \le \theta \le 90^\circ$, the maximum value of $\sin\theta$ is $1$ (when $\theta = 90^\circ$).
Answer: d) $1$
- $ABC$ is a right-angled triangle whose hypotenuse $AC = 2$, $AB = 1$ such that —
(i) $\angle ACB = 30^\circ$
(ii) $\tan A = \sqrt{3}$
(iii) $\sin(A + C) = 0$
Solution:
$\sin C = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{1}{2} \implies \sin C = \sin 30^\circ \implies \angle ACB = 30^\circ$ [Statement (i) is true]
Since $\angle B = 90^\circ$ and $C = 30^\circ$, therefore $A = 90^\circ – 30^\circ = 60^\circ$
$\tan A = \tan 60^\circ = \sqrt{3}$ [Statement (ii) is true]
Again, $A + C = 60^\circ + 30^\circ = 90^\circ$
$\sin(A + C) = \sin 90^\circ = 1 \neq 0$ [Statement (iii) is false]
Answer: c) i and ii
For questions 3 and 4:
In right-angled triangle $ABC$, hypotenuse $AC = 2$, $AB = 1$.
By Pythagoras’ theorem, $BC = \sqrt{AC^2 – AB^2} = \sqrt{2^2 – 1^2} = \sqrt{3}$
- What is the value of $\tan C$?
Solution:
$\tan C = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} = \frac{1}{\sqrt{3}}$
Answer: d) $\frac{1}{\sqrt{3}}$
- What is the value of $\sin^2 A – \cos^2 A$?
Solution:
With respect to angle $A$, Perpendicular $= BC = \sqrt{3}$, Base $= AB = 1$, Hypotenuse $= AC = 2$
$\sin A = \frac{\sqrt{3}}{2}$ and $\cos A = \frac{1}{2}$
Given Expression $= \sin^2 A – \cos^2 A$
$= \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{1}{2}\right)^2$
$= \frac{3}{4} – \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
Answer: a) $\frac{1}{2}$
- Creative Question:
$\sin\theta = p$, $\cos\theta = q$, $\tan\theta = r$, where $\theta$ is an acute angle.
(a) If $r = \sqrt{(3)^{-1}}$, find the value of $\theta$.
Solution:
Given, $r = \sqrt{(3)^{-1}}$
or, $\tan\theta = \sqrt{\frac{1}{3}}$
or, $\tan\theta = \frac{1}{\sqrt{3}}$
or, $\tan\theta = \tan 30^\circ$
or, $\theta = 30^\circ$
Answer: $\theta = 30^\circ$
(b) If $p + q = \sqrt{2}$, prove that $\theta = 45^\circ$.
Proof:
Given, $p + q = \sqrt{2}$
or, $\sin\theta + \cos\theta = \sqrt{2}$
or, $\sin\theta = \sqrt{2} – \cos\theta$
[Squaring both sides]
or, $\sin^2\theta = (\sqrt{2} – \cos\theta)^2$
or, $1 – \cos^2\theta = 2 – 2\sqrt{2}\cos\theta + \cos^2\theta$
or, $2\cos^2\theta – 2\sqrt{2}\cos\theta + 1 = 0$
or, $(\sqrt{2}\cos\theta – 1)^2 = 0$
or, $\sqrt{2}\cos\theta – 1 = 0$
or, $\cos\theta = \frac{1}{\sqrt{2}}$
or, $\cos\theta = \cos 45^\circ$
or, $\theta = 45^\circ$ (Proved)
(c) If $7p^2 + 3q^2 = 4$, show that $\tan\theta = \frac{1}{\sqrt{3}}$.
Proof:
Given, $7p^2 + 3q^2 = 4$
or, $7\sin^2\theta + 3\cos^2\theta = 4$
or, $4\sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = 4$
or, $4\sin^2\theta + 3(\sin^2\theta + \cos^2\theta) = 4$
or, $4\sin^2\theta + 3(1) = 4$
or, $4\sin^2\theta = 1$
or, $\sin^2\theta = \frac{1}{4}$
or, $\sin\theta = \frac{1}{2}$ ($\because \theta$ is an acute angle)
or, $\sin\theta = \sin 30^\circ \implies \theta = 30^\circ$
Now, $\tan\theta = \tan 30^\circ = \frac{1}{\sqrt{3}}$ (Shown)
- Short-Answer Questions:
(a) Show that $\tan^2 A – \sin^2 A = \tan^2 A \cdot \sin^2 A$
Solution:
L.H.S. $= \tan^2 A – \sin^2 A$
$= \tan^2 A \left(1 – \frac{\sin^2 A}{\tan^2 A}\right)$
$= \tan^2 A \left(1 – \frac{\sin^2 A}{\frac{\sin^2 A}{\cos^2 A}}\right)$
$= \tan^2 A (1 – \cos^2 A)$
$= \tan^2 A \cdot \sin^2 A$
$= \text{R.H.S.}$ (Shown)
(b) In triangle $\triangle ABC$, if $\angle B = 90^\circ$, $BC = 6\text{ cm}$, and $AB = 8\text{ cm}$, then find the value of $\sin A$.
Solution:
According to Pythagoras’ theorem,
$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}$
With respect to angle $A$, Perpendicular $= BC = 6\text{ cm}$ and Hypotenuse $= AC = 10\text{ cm}$
$\sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{6}{10} = \frac{3}{5}$
Answer: $\sin A = \frac{3}{5}$
(c) If $\sin^2 A = 1 + \cos^2 A$, find the value of $A$ (assuming $A$ is an acute angle).
Solution:
$\sin^2 A = 1 + \cos^2 A$
or, $1 – \cos^2 A = 1 + \cos^2 A$
or, $2\cos^2 A = 0$
or, $\cos A = 0$
or, $\cos A = \cos 90^\circ \implies A = 90^\circ$
Since $A$ is an acute angle ($0^\circ < A < 90^\circ$), $A = 90^\circ$ is not acceptable. Therefore, according to the given condition, there is no real value for the acute angle $A$.
Answer: Value is not acceptable / No solution.
(d) If $A = 15^\circ$, then show that $\cos^3 3A = \sin^3 3A$.
Proof:
Given, $A = 15^\circ$
Therefore, $3A = 3 \times 15^\circ = 45^\circ$
L.H.S. $= \cos^3 3A = \cos^3 45^\circ$
$= \left(\frac{1}{\sqrt{2}}\right)^3$
$= \frac{1}{2\sqrt{2}}$
R.H.S. $= \sin^3 3A = \sin^3 45^\circ$
$= \left(\frac{1}{\sqrt{2}}\right)^3$
$= \frac{1}{2\sqrt{2}}$
Therefore, $\text{L.H.S.} = \text{R.H.S.}$ (Shown)