SEBA Class 10 Advanced Maths Chapter 7.1 Solutions: Trigonometry
Looking for complete and accurate SEBA Class 10 Advanced Maths Chapter 7.1 Solutions for your upcoming HSLC board examination? In this article, we provide step-by-step, easy-to-understand solutions to all textual exercises of Chapter 7: Trigonometry from the SEBA Class 10 New Advanced Mathematics textbook.
1. Establish the signs of the following trigonometrical ratios:
(i) $\sin 77^\circ$
$77^\circ$ is in the first quadrant and all trigonometrical ratios of angles in the first quadrant are positive. Therefore, $\sin 77^\circ$ is positive.
(ii) $\cos 125^\circ$
$125^\circ$ is in the second quadrant in which only ‘$\sin$‘ and ‘$\operatorname{cosec}$‘ are positive while all of the other ratios are negative.
Thus $\cos 125^\circ$ is negative.
(iii) $\tan 714^\circ$
$714^\circ = 7 \times 90^\circ + 84^\circ$ is an angle in the fourth quadrant in which ‘$\cos$‘ and ‘$\sec$‘ are positive while the other trigonometrical ratios are negative. Therefore, $\tan 714^\circ$ is negative.
(iv) $\sec 1510^\circ$
$1510^\circ = 16 \times 90^\circ + 70^\circ$ is an angle in the first quadrant and all trigonometrical ratios of angles in the first quadrant are positive.
Therefore, $\sec 1510^\circ$ is positive.
(v) $\sin 2013^\circ$
$2013^\circ = 22 \times 90^\circ + 33^\circ$ is an angle in the third quadrant in which only ‘$\tan$‘ and ‘$\cot$‘ are positive while all of the other trigonometrical ratios are negative.
Thus $\sin 2013^\circ$ is negative.
(vi) $\cot(-777^\circ)$
We have $\cot(-777^\circ) = -\cot 777^\circ$. Now, $777^\circ = 8 \times 90^\circ + 57^\circ$ is an angle in the first quadrant in which all trigonometrical ratios are positive, so $\cot 777^\circ$ is positive.
Therefore, $-\cot 777^\circ$ is negative, which means $\cot(-777^\circ)$ is negative.
(vii) $\operatorname{cosec}(-1208^\circ)$
We have $\operatorname{cosec}(-1208^\circ) = -\operatorname{cosec} 1208^\circ$. Now, $1208^\circ = 13 \times 90^\circ + 38^\circ$ is an angle in the second quadrant in which ‘$\sin$’ and ‘$\operatorname{cosec}$’ are positive, so $\operatorname{cosec} 1208^\circ$ is positive.
Therefore, $-\operatorname{cosec} 1208^\circ$ is negative, which means $\operatorname{cosec}(-1208^\circ)$ is negative.
(viii) $\cos(-1895^\circ)$
We have $\cos(-1895^\circ) = \cos 1895^\circ$. Now, $1895^\circ = 21 \times 90^\circ + 5^\circ$ is an angle in the second quadrant in which only ‘$\sin$’ and ‘$\operatorname{cosec}$’ are positive while all of the other trigonometrical ratios are negative.
Therefore, $\cos(-1895^\circ)$ is negative.
2. Express the following trigonometrical ratios in terms of trigonometrical ratios of positive acute angles:
(i) $\sin 130^\circ$
$= \sin(1 \times 90^\circ + 40^\circ)$
$= \cos 40^\circ $
Since $n = 1$ (odd), $\sin$ changes to $\cos$.
The angle $130^\circ$ lies in the 2nd quadrant, where $\sin$ is positive.]
(ii) $\sec 190^\circ$
$= \sec(2 \times 90^\circ + 10^\circ)$
$= -\sec 10^\circ$
[Explanation:-
Since $n = 2$ (even), the ratio remains $\sec$.
The angle $190^\circ$ lies in the 3rd quadrant, where $\sec$ is negative.]
(iii) $\cot 200^\circ$
$= \cot(2 \times 90^\circ + 20^\circ)$
$= \cot 20^\circ$
[Explanation:-
Since $n = 2$ (even), the ratio remains $\cot$.
The angle $200^\circ$ lies in the 3rd quadrant, where $\cot$ is positive.]
(iv) $\cos 300^\circ$
$= \cos(3 \times 90^\circ + 30^\circ)$
$= \sin 30^\circ$
[Explanation:-
Since $n = 3$ (odd), $\cos$ changes to $\sin$.
The angle $300^\circ$ lies in the 4th quadrant, where $\cos$ is positive.]
(v) $\operatorname{cosec} 420^\circ$
$= \operatorname{cosec}(4 \times 90^\circ + 60^\circ)$
$= \operatorname{cosec} 60^\circ$
[Explanation:-
Since $n = 4$ (even), the ratio remains $\operatorname{cosec}$.
The angle $420^\circ$ lies in the 1st quadrant, where $\operatorname{cosec}$ is positive.]
(vi) $\sin 720^\circ$
$= \sin(8 \times 90^\circ + 0^\circ)$
$= \sin 0^\circ$
[Explanation:-
Since $n = 8$ (even), the ratio remains $\sin$.
The terminal ray coincides with the positive x-axis.]
(vii) $\tan 620^\circ$
$= \tan(6 \times 90^\circ + 80^\circ)$
$= \tan 80^\circ$
[Explanation:-
Since $n = 6$ (even), the ratio remains $\tan$.
The angle $620^\circ$ lies in the 3rd quadrant, where $\tan$ is positive.]
(viii) $\cos 1000^\circ$
$= \cos(11 \times 90^\circ + 10^\circ)$
$= \sin 10^\circ$
[Explanation:-
Since $n = 11$ (odd), $\cos$ changes to $\sin$.
The angle $1000^\circ$ is lies in the 4th quadrant where cosine is strictly positive ($+\sin 10^\circ$).
(ix) $\tan(-320^\circ)$
$= -\tan 320^\circ$
$= -\tan(3 \times 90^\circ + 50^\circ)$
$= -\cot 50^\circ$
[Explanation:-
We know $\tan(-\theta) = -\tan \theta$.
For $\tan(3 \times 90^\circ + 50^\circ)$, since $n = 3$ (odd), $\tan$ changes to $\cot$.]
(x) $\sec(-945^\circ)$
$= \sec 945^\circ$
$= \sec(10 \times 90^\circ + 45^\circ)$
$= -\sec 45^\circ$
[Explanation:-
We know $\sec(-\theta) = \sec \theta$.
For $\sec(10 \times 90^\circ + 45^\circ)$, since $n = 10$ (even), the ratio remains $\sec$.
The angle $945^\circ$ lies in the 3rd quadrant, where $\sec$ is negative.]
3. Evaluate:
(i) $\sin^3 120^\circ$
$= \sin^3(90^\circ + 30^\circ)$
$= \cos^3 30^\circ$
$= \left(\frac{\sqrt{3}}{2}\right)^3$
$= \frac{3\sqrt{3}}{8}$
(ii) $\cos^2 135^\circ$
$= \cos^2(90^\circ + 45^\circ)$
$= (-\sin 45^\circ)^2$
$= \sin^2 45^\circ$
$= \left(\frac{1}{\sqrt{2}}\right)^2$
$= \frac{1}{2}$
(iii) $\tan^2 330^\circ$
$= \tan^2(3 \times 90^\circ + 60^\circ)$
$= (-\cot 60^\circ)^2$
$= \cot^2 60^\circ$
$= \left(\frac{1}{\sqrt{3}}\right)^2$
$= \frac{1}{3}$
(iv) $\operatorname{cosec}^3 300^\circ + \sec^3 300^\circ$
$= \operatorname{cosec}^3(3 \times 90^\circ + 30^\circ) + \sec^3(3 \times 90^\circ + 30^\circ)$
$= (-\sec 30^\circ)^3 + (\operatorname{cosec} 30^\circ)^3$
$= \left(-\frac{2}{\sqrt{3}}\right)^3 + (2)^3$
$= -\frac{8}{3\sqrt{3}} + 8$
$= \frac{24\sqrt{3} – 8}{3\sqrt{3}}$
$= \frac{8(\sqrt{3} – 1)}{3\sqrt{3}}$
(v) $\sin^2 330^\circ + \cos^2 30^\circ$
$= \sin^2(3 \times 90^\circ + 60^\circ) + \cos^2 30^\circ$
$= (-\cos 60^\circ)^2 + \cos^2 30^\circ$
$= \cos^2 60^\circ + \cos^2 30^\circ$
$= \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$
$= \frac{1}{4} + \frac{3}{4}$
$= \frac{4}{4}$
$= 1$
(vi) $\cos 63^\circ \cos 27^\circ + \sin 63^\circ \sin 27^\circ$
$= \cos(90^\circ – 27^\circ) \cos 27^\circ + \sin(90^\circ – 27^\circ) \sin 27^\circ$
$= \sin 27^\circ \cos 27^\circ + \cos 27^\circ \sin 27^\circ$
$= 2 \sin 27^\circ \cos 27^\circ$
$= \sin(2 \times 27^\circ)$
(vii) $\sin^2 210^\circ + \cos^2 30^\circ – \cot^2 45^\circ$
$= \sin^2(2 \times 90^\circ + 30^\circ) + \cos^2 30^\circ – \cot^2 45^\circ$
$= (-\sin 30^\circ)^2 + \cos^2 30^\circ – \cot^2 45^\circ$
$= \sin^2 30^\circ + \cos^2 30^\circ – \cot^2 45^\circ$
$= 1 – (1)^2$
$= 1 – 1$
$= 0$
(viii) $\frac{\sin 300^\circ}{\tan 135^\circ} \cdot \frac{\tan 330^\circ}{\sin 210^\circ} \cdot \frac{\sec 420^\circ}{\sec 315^\circ}$
$= \frac{\sin(3 \times 90^\circ + 30^\circ)}{\tan(90^\circ + 45^\circ)} \cdot \frac{\tan(3 \times 90^\circ + 60^\circ)}{\sin(2 \times 90^\circ + 30^\circ)} \cdot \frac{\sec(4 \times 90^\circ + 60^\circ)}{\sec(3 \times 90^\circ + 45^\circ)}$
$= \frac{-\cos 30^\circ}{-\cot 45^\circ} \cdot \frac{-\cot 60^\circ}{-\sin 30^\circ} \cdot \frac{\sec 60^\circ}{\operatorname{cosec} 45^\circ}$
$= \frac{\cos 30^\circ}{\cot 45^\circ} \cdot \frac{\cot 60^\circ}{\sin 30^\circ} \cdot \frac{\sec 60^\circ}{\operatorname{cosec} 45^\circ}$
$= \frac{\frac{\sqrt{3}}{2}}{1} \cdot \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}} \cdot \frac{2}{\sqrt{2}}$
$= \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{2}{\sqrt{3}}\right) \cdot \left(\frac{2}{\sqrt{2}}\right)$
$= 1 \cdot \sqrt{2}$
$= \sqrt{2}$
(ix) $\frac{\tan 495^\circ}{\cot 855^\circ}$
$= \frac{\tan(5 \times 90^\circ + 45^\circ)}{\cot(9 \times 90^\circ + 45^\circ)}$
$= \frac{-\cot 45^\circ}{-\tan 45^\circ}$
$= \frac{-1}{-1}$
$= 1$
4. Show that:
(i) $\sin(90^\circ – A)\sin(180^\circ + A) + \cos(90^\circ + A)\cos(180^\circ – A) = 0$
$\text{L.H.S.} = \sin(90^\circ – A)\sin(180^\circ + A) + \cos(90^\circ + A)\cos(180^\circ – A)$
$= \cos A \cdot (-\sin A) + (-\sin A) \cdot (-\cos A)$
$= -\sin A \cos A + \sin A \cos A$
$= 0$
$= \text{R.H.S.}$
Showed.
(ii) $\sin A + \cos(90^\circ + A) + \tan(180^\circ + A) + \cot(270^\circ + A) = 0$
$\text{L.H.S.} = \sin A + \cos(90^\circ + A) + \tan(180^\circ + A) + \cot(270^\circ + A)$
$= \sin A + (-\sin A) + \tan A + (-\tan A)$
$= \sin A – \sin A + \tan A – \tan A$
$= 0$
$= \text{R.H.S.}$
Showed.
(iii) $\sin 420^\circ \cos 390^\circ + \cos(-300^\circ)\sin(-330^\circ) = 1$
$\text{L.H.S.} = \sin 420^\circ \cos 390^\circ + \cos(-300^\circ)\sin(-330^\circ)$
$= \sin 420^\circ \cos 390^\circ – \cos 300^\circ \sin 330^\circ \quad [\because \cos(-\theta) = \cos\theta, \sin(-\theta) = -\sin\theta]$
$= \sin(4 \times 90^\circ + 60^\circ) \cos(4 \times 90^\circ + 30^\circ) – \cos(3 \times 90^\circ + 30^\circ)\sin(3 \times 90^\circ + 60^\circ)$
$= \sin 60^\circ \cos 30^\circ – \sin 30^\circ(-\cos 60^\circ)$
$= \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2}$
$= \frac{3}{4} + \frac{1}{4}$
$= \frac{3+1}{4}$
$= \frac{4}{4}$
$= 1$
$= \text{R.H.S.}$
Showed.
(iv) $\sin^2\frac{\pi}{3} + \sin^2\frac{2\pi}{3} + \sin^2\frac{5\pi}{3} + \sin^2\frac{8\pi}{3} = 3 \quad [\pi \text{ radian} = 180^\circ]$
$\text{L.H.S.} = \sin^2\frac{\pi}{3} + \sin^2\frac{2\pi}{3} + \sin^2\frac{5\pi}{3} + \sin^2\frac{8\pi}{3}$
$= \sin^2\frac{\pi}{3} + \sin^2\left(\pi – \frac{\pi}{3}\right) + \sin^2\left(2\pi – \frac{\pi}{3}\right) + \sin^2\left(3\pi – \frac{\pi}{3}\right)$
$= \sin^2\frac{\pi}{3} + \sin^2\frac{\pi}{3} + \left(-\sin\frac{\pi}{3}\right)^2 + \sin^2\frac{\pi}{3}$
$= \sin^2\frac{\pi}{3} + \sin^2\frac{\pi}{3} + \sin^2\frac{\pi}{3} + \sin^2\frac{\pi}{3}$
$= 4\sin^2\frac{\pi}{3}$
$= 4 \times \left(\frac{\sqrt{3}}{2}\right)^2$
$= 4 \times \frac{3}{4}$
$= 3$
$= \text{R.H.S.}$
Showed.
(v) $\frac{\sin 33^\circ}{\cos 57^\circ} – \frac{\sin 23^\circ}{\cos 67^\circ} = 0$
$\text{L.H.S.} = \frac{\sin 33^\circ}{\cos 57^\circ} – \frac{\sin 23^\circ}{\cos 67^\circ}$
$= \frac{\sin(90^\circ – 57^\circ)}{\cos 57^\circ} – \frac{\sin(90^\circ – 67^\circ)}{\cos 67^\circ}$
$= \frac{\cos 57^\circ}{\cos 57^\circ} – \frac{\cos 67^\circ}{\cos 67^\circ}$
$= 1 – 1$
$= 0$
$= \text{R.H.S.}$
Showed.
(vi) $\frac{\sin^2 42^\circ + \sin^2 48^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} = 1$
$\text{L.H.S.} = \frac{\sin^2 42^\circ + \sin^2 48^\circ}{\cos^2 24^\circ + \cos^2 66^\circ}$
$= \frac{\sin^2(90^\circ – 48^\circ) + \sin^2 48^\circ}{\cos^2(90^\circ – 66^\circ) + \cos^2 66^\circ}$
$= \frac{\cos^2 48^\circ + \sin^2 48^\circ}{\sin^2 66^\circ + \cos^2 66^\circ}$
$= \frac{1}{1}$
$= 1$
$= \text{R.H.S.}$
Showed.
(vii) $\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \dots \tan 88^\circ \cdot \tan 89^\circ = 1$
$\text{L.H.S.} = \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \dots \tan 88^\circ \cdot \tan 89^\circ$
$= (\tan 1^\circ \cdot \tan 89^\circ) \cdot (\tan 2^\circ \cdot \tan 88^\circ) \dots (\tan 44^\circ \cdot \tan 46^\circ) \cdot \tan 45^\circ$
$= \{\tan(90^\circ – 89^\circ)\tan 89^\circ\} \cdot \{\tan(90^\circ – 88^\circ)\tan 88^\circ\} \dots \tan 45^\circ$
$= (\cot 89^\circ \cdot \tan 89^\circ) \cdot (\cot 88^\circ \cdot \tan 88^\circ) \dots 1$
$= 1 \cdot 1 \dots 1$
$= 1$
$= \text{R.H.S.}$
Showed.
(viii) $\sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 90^\circ = 9\frac{1}{2}$
$\text{L.H.S.} = \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2 90^\circ$
$= \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \sin^2(90^\circ – 15^\circ) + \sin^2(90^\circ – 10^\circ) + \sin^2(90^\circ – 5^\circ) + \sin^2 90^\circ$
$= \sin^2 5^\circ + \sin^2 10^\circ + \sin^2 15^\circ + \dots + \cos^2 15^\circ + \cos^2 10^\circ + \cos^2 5^\circ + 1^2$
$= (\sin^2 5^\circ + \cos^2 5^\circ) + (\sin^2 10^\circ + \cos^2 10^\circ) + \dots + (\sin^2 40^\circ + \cos^2 40^\circ) + \sin^2 45^\circ + \sin^2 90^\circ$
$= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + \left(\frac{1}{\sqrt{2}}\right)^2 + 1$
$= 8 + \frac{1}{2} + 1$
$= 9 + \frac{1}{2}$
$= 9\frac{1}{2}$
$= \text{R.H.S.}$
Showed.
5. (a) Show that the following expression is independent of the value of $A$:
$\frac{\sin^3(180^\circ + A)\cos^3(-A)\tan(360^\circ – A)\sec^2(180^\circ – A)}{\cos^2(90^\circ + A)\sin^3(180^\circ – A)\operatorname{cosec}^2 A \sin(360^\circ + A)}$
Soln :
$\text{L.H.S.} = \frac{\sin^3(180^\circ + A)\cos^3(-A)\tan(360^\circ – A)\sec^2(180^\circ – A)}{\cos^2(90^\circ + A)\sin^3(180^\circ – A)\operatorname{cosec}^2 A \sin(360^\circ + A)}$
$= \frac{(-\sin A)^3 (\cos A)^3 (-\tan A)(-\sec A)^2}{(-\sin A)^2 (\sin A)^3 \operatorname{cosec}^2 A (\sin A)}$
$= \frac{-\sin^3 A \cdot \cos^3 A \cdot (-\tan A) \cdot \sec^2 A}{\sin^2 A \cdot \sin^3 A \cdot \operatorname{cosec}^2 A \cdot \sin A}$
$= \frac{\sin^3 A \cdot \cos^3 A \cdot \tan A \cdot \sec^2 A}{\sin^6 A \cdot \operatorname{cosec}^2 A}$
$= \frac{\cos^3 A \cdot \left(\frac{\sin A}{\cos A}\right) \cdot \left(\frac{1}{\cos^2 A}\right)}{\sin^3 A \cdot \left(\frac{1}{\sin^2 A}\right)}$
$= \frac{\sin A}{\sin A}$
$= 1$
Which is independent of the value of $A$.
Showed.
5. (b) If $\tan 25^\circ = t$, then express the following in terms of $t$:
(i) $\frac{\tan 205^\circ – \tan 115^\circ}{\tan 245^\circ + \tan 335^\circ}$
Soln :
$ = \frac{\tan 205^\circ – \tan 115^\circ}{\tan 245^\circ + \tan 335^\circ}$
$= \frac{\tan(180^\circ + 25^\circ) – \tan(90^\circ + 25^\circ)}{\tan(270^\circ – 25^\circ) + \tan(360^\circ – 25^\circ)}$
$= \frac{\tan 25^\circ – (-\cot 25^\circ)}{\cot 25^\circ + (-\tan 25^\circ)}$
$= \frac{\tan 25^\circ + \cot 25^\circ}{\cot 25^\circ – \tan 25^\circ}$
$= \frac{t + \frac{1}{t}}{\frac{1}{t} – t}$
$= \frac{\frac{t^2 + 1}{t}}{\frac{1 – t^2}{t}}$
$= \frac{1 + t^2}{1 – t^2}$
(ii) $\frac{\tan 155^\circ – \tan 115^\circ}{1 + \tan 155^\circ \tan 115^\circ}$
Soln :
$\frac{\tan 155^\circ – \tan 115^\circ}{1 + \tan 155^\circ \tan 115^\circ}$
$= \frac{\tan(180^\circ – 25^\circ) – \tan(90^\circ + 25^\circ)}{1 + \tan(180^\circ – 25^\circ)\tan(90^\circ + 25^\circ)}$
$= \frac{-\tan 25^\circ – (-\cot 25^\circ)}{1 + (-\tan 25^\circ)(-\cot 25^\circ)}$
$= \frac{-\tan 25^\circ + \cot 25^\circ}{1 + \tan 25^\circ \cot 25^\circ}$
$= \frac{-t + \frac{1}{t}}{1 + 1} \quad [\because \tan 25^\circ \cdot \cot 25^\circ = 1]$
$= \frac{\frac{-t^2 + 1}{t}}{2}$
$= \frac{1 – t^2}{2t}$
6. Applying the formulae for trigonometrical ratios of associated angles, find the value of $\theta$ in the given domains:
(i) $\sin\theta = \frac{1}{2} \quad (0^\circ < \theta < 360^\circ)$
Soln :
$\sin\theta = \frac{1}{2} = \sin 30^\circ$
$\therefore \theta = 30^\circ$
Again, since $\sin\theta$ is also positive in the 2nd quadrant:
$\sin\theta = \sin 30^\circ$
$= \sin(180^\circ – 30^\circ)$
$= \sin 150^\circ$
$\therefore \theta = 150^\circ$
$\therefore \theta = 30^\circ, 150^\circ$
(ii) $\cos\theta = \frac{1}{\sqrt{2}} \quad (0^\circ < \theta < 360^\circ)$
Soln :
$\cos\theta = \frac{1}{\sqrt{2}} = \cos 45^\circ$
$\therefore \theta = 45^\circ$
Again, since $\cos\theta$ is also positive in the 4th quadrant:
$\cos\theta = \cos 45^\circ$
$= \cos(360^\circ – 45^\circ)$
$= \cos(4 \times 90^\circ – 45^\circ)$
$= \cos 315^\circ$
$\therefore \theta = 315^\circ$
$\therefore \theta = 45^\circ, 315^\circ$
(iii) $\tan\theta = -\sqrt{3} \quad (0^\circ < \theta < 360^\circ)$
Soln :
$\tan\theta = -\sqrt{3}$
$\tan\theta = -\tan 60^\circ$
Since $\tan\theta$ is negative in the 2nd and 4th quadrants:
In the 2nd quadrant:
$\tan\theta = \tan(180^\circ – 60^\circ)$
$\implies \tan\theta = \tan 120^\circ$ — (i)
In the 4th quadrant:
$\tan\theta = \tan(360^\circ – 60^\circ)$
$\implies \tan\theta = \tan 300^\circ$ — (ii)
From (i) and (ii), we have:
$\theta = 120^\circ, 300^\circ$
(iv) $2\sin^2\theta + 3\sin\theta + 1 = 0 \quad (180^\circ < \theta < 540^\circ)$
Soln :
$2\sin^2\theta + 3\sin\theta + 1 = 0$
$\implies 2\sin^2\theta + 2\sin\theta + \sin\theta + 1 = 0$
$\implies 2\sin\theta(\sin\theta + 1) + 1(\sin\theta + 1) = 0$
$\implies (\sin\theta + 1)(2\sin\theta + 1) = 0$
Either,
$\sin\theta + 1 = 0$
$\implies \sin\theta = -1$
$\implies \sin\theta = -\sin 90^\circ$
$\implies \sin\theta= \sin(180^\circ + 90^\circ)$
$\implies \sin\theta= \sin 270^\circ$
$\therefore \theta = 270^\circ$ — (i)
Or,
$2\sin\theta + 1 = 0$
$\implies \sin\theta = -\frac{1}{2}$
$\implies \sin\theta = -\sin 30^\circ$
$\implies \sin\theta = \sin(180^\circ + 30^\circ)$
$\implies \sin\theta = \sin 210^\circ$
$\therefore \theta = 210^\circ$ — (ii)
Also, for the 4th quadrant:
$\sin\theta = \sin(360^\circ – 30^\circ)$
$\sin\theta = \sin 330^\circ$
$\therefore \theta = 330^\circ$ — (iii)
From (i), (ii), and (iii), we have:
$\theta = 210^\circ, 270^\circ, 330^\circ$
(v) $2\sin^2\theta + \sqrt{3}\cos\theta + 1 = 0 \quad (90^\circ < \theta < 270^\circ)$
Soln :
$2\sin^2\theta + \sqrt{3}\cos\theta + 1 = 0$
$\implies 2(1 – \cos^2\theta) + \sqrt{3}\cos\theta + 1 = 0$
$\implies 2 – 2\cos^2\theta + \sqrt{3}\cos\theta + 1 = 0$
$\implies -2\cos^2\theta + \sqrt{3}\cos\theta + 3 = 0$
$\implies 2\cos^2\theta – \sqrt{3}\cos\theta – 3 = 0$
$\implies 2\cos^2\theta – 2\sqrt{3}\cos\theta + \sqrt{3}\cos\theta – 3 = 0$
$\implies 2\cos\theta(\cos\theta – \sqrt{3}) + \sqrt{3}(\cos\theta – \sqrt{3}) = 0$
$\implies (\cos\theta – \sqrt{3})(2\cos\theta + \sqrt{3}) = 0$
Either,
$\cos\theta – \sqrt{3} = 0$
$\implies \cos\theta = \sqrt{3}$ (Not acceptable as $-1 \le \cos\theta \le 1$)
Or,
$2\cos\theta + \sqrt{3} = 0$
$\implies \cos\theta = -\frac{\sqrt{3}}{2}$
$\implies \cos\theta = -\cos 30^\circ$
In the 2nd quadrant ($90^\circ < \theta < 180^\circ$):
$\cos\theta = \cos(180^\circ – 30^\circ)$
$\implies \cos\theta = \cos 150^\circ$ — (i)
$\therefore \theta = 150^\circ$
In the 3rd quadrant ($180^\circ < \theta < 270^\circ$):
$\cos\theta = \cos(180^\circ + 30^\circ)$
$\implies \cos\theta = \cos 210^\circ$ — (ii)
$\therefore \theta = 210^\circ$
From (i) and (ii):
$\theta = 150^\circ, 210^\circ$
(v): Solve $2\sin^2\theta + \sqrt{3}\cos\theta + 1 = 0 \quad (90^\circ < \theta < 270^\circ)$
Soln :
$2\sin^2\theta + \sqrt{3}\cos\theta + 1 = 0$
$\implies 2(1 – \cos^2\theta) + \sqrt{3}\cos\theta + 1 = 0$ [ Since, $\sin^2\theta = 1 – \cos^2\theta$]
$\implies 2 – 2\cos^2\theta + \sqrt{3}\cos\theta + 1 = 0$
$\implies -2\cos^2\theta + \sqrt{3}\cos\theta + 3 = 0$
$\implies 2\cos^2\theta – \sqrt{3}\cos\theta – 3 = 0$
$\implies 2x^2 – \sqrt{3}x – 3 = 0$ [Let $\cos\theta = x$]
$\implies 2x^2 – (2\sqrt{3} + \sqrt{3})x – 3 = 0$
$\implies 2x^2 – 2\sqrt{3}x + \sqrt{3}x – 3 = 0$
$\implies 2x(x – \sqrt{3}) + \sqrt{3}(x – \sqrt{3}) = 0$
$\implies (x – \sqrt{3})(2x + \sqrt{3}) = 0$
Either,
$x – \sqrt{3} = 0$
$\implies x = \sqrt{3}$
$\implies \cos\theta = \sqrt{3}$
(Not acceptable, since $-1 \le \cos\theta \le 1$)
Or,
$2x + \sqrt{3} = 0$
$\implies 2x = -\sqrt{3}$
$\implies x = -\frac{\sqrt{3}}{2}$
$\implies \cos\theta = -\frac{\sqrt{3}}{2}$
$\implies \cos\theta = -\cos 30^\circ$
Since $\cos\theta$ is negative in the 2nd and 3rd quadrants and the given domain is $90^\circ < \theta < 270^\circ$:
In the 2nd quadrant:
$\cos\theta = \cos(180^\circ – 30^\circ)$
$\implies \cos\theta = \cos 150^\circ$ — (i)
$\therefore \theta = 150^\circ$
In the 3rd quadrant:
$\cos\theta = \cos(180^\circ + 30^\circ)$
$\implies \cos\theta = \cos 210^\circ$ — (ii)
$\therefore \theta = 210^\circ$
From (i) and (ii), we get:
$\theta = 150^\circ, 210^\circ$
7.(a) For the angles $A, B, C$ of a triangle, show that: (i) $\tan(A+B) + \tan C = 0$
Soln :
Since $A, B, C$ are the angles of a triangle:
$A + B + C = 180^\circ \quad [\because \text{angle sum property of a triangle}]$
$\implies A + B = 180^\circ – C$
Taking tangent on both sides:
$\implies \tan(A + B) = \tan(180^\circ – C)$
$\implies \tan(A + B) = -\tan C$
$\therefore \tan(A + B) + \tan C = 0$ Proved.
(ii) $\frac{\sin(B+C) + \sin(C+A) + \sin(A+B)}{\sin A + \sin B + \sin C} = 1$
Soln :
Since $A, B, C$ are the angles of a triangle:
$A + B + C = 180^\circ$
$\implies B + C = 180^\circ – A$
$\implies \sin(B + C) = \sin(180^\circ – A) = \sin A$
Similarly,
$C + A = 180^\circ – B \implies \sin(C + A) = \sin(180^\circ – B) = \sin B$
$A + B = 180^\circ – C \implies \sin(A + B) = \sin(180^\circ – C) = \sin C$
Now,
$\text{L.H.S.} = \frac{\sin(B + C) + \sin(C + A) + \sin(A + B)}{\sin A + \sin B + \sin C}$
$= \frac{\sin A + \sin B + \sin C}{\sin A + \sin B + \sin C}$
$= 1$
$= \text{R.H.S.}$ Proved.
(iii) $\frac{\tan(\pi – A) + \tan(2\pi – A) + \tan(3\pi – A)}{\tan(4\pi – A) + \tan(5\pi – A) + \tan(6\pi – A)} = 1$
Soln :
Since $A, B, C$ are the angles of a triangle, $A + B + C = \pi$ ($180^\circ$).
We know that for any integer $k$:
$\tan(k\pi – A) = -\tan A$
Therefore:
- $\tan(\pi – A) = -\tan A$
- $\tan(2\pi – A) = -\tan A$
- $\tan(3\pi – A) = -\tan A$
- $\tan(4\pi – A) = -\tan A$
- $\tan(5\pi – A) = -\tan A$
- $\tan(6\pi – A) = -\tan A$
$\text{L.H.S.} = \frac{\tan(\pi – A) + \tan(2\pi – A) + \tan(3\pi – A)}{\tan(4\pi – A) + \tan(5\pi – A) + \tan(6\pi – A)}$
$= \frac{-\tan A – \tan A – \tan A}{-\tan A – \tan A – \tan A}$
$= \frac{-3\tan A}{-3\tan A}$
$= 1$
$= \text{R.H.S.}$
Proved.
7.(b)(i) If $A, B, C$ and $D$ are angles of a quadrilateral, then show that:
$\cos\frac{A+B}{2} + \cos\frac{C+D}{2} = 0$
$Soln :
Since $A, B, C, D$ are the four angles of a quadrilateral:
$A + B + C + D = 360^\circ$
$\implies A + B = 360^\circ – (C + D)$
$\implies \frac{A + B}{2} = 180^\circ – \frac{C + D}{2}$
$\implies \cos\left(\frac{A + B}{2}\right) = \cos\left(180^\circ – \frac{C + D}{2}\right)$ [ Taking cosine on both sides]
$\implies \cos\left(\frac{A + B}{2}\right) = -\cos\left(\frac{C + D}{2}\right)$
$\therefore \cos\left(\frac{A + B}{2}\right) + \cos\left(\frac{C + D}{2}\right) = 0$ Proved.
(ii) If $A, B, C$ and $D$ are angles of a cyclic quadrilateral, show that:
$\cos A + \cos B + \cos C + \cos D = 0$
$Soln :
Since $A, B, C, D$ are the angles of a cyclic quadrilateral, the opposite angles are supplementary:
$A + C = 180^\circ$
$\implies A = 180^\circ – C$
$\therefore \cos A = \cos(180^\circ – C) = -\cos C$ — (i) Again,
$B + D = 180^\circ$
$\implies B = 180^\circ – D$
$\therefore \cos B = \cos(180^\circ – D) = -\cos D$ — (ii)
Now,
$\text{L.H.S.} = \cos A + \cos B + \cos C + \cos D$
$= (-\cos C) + (-\cos D) + \cos C + \cos D \quad [\text{using (i) and (ii)}]$
$= -\cos C + \cos C – \cos D + \cos D$
$= 0$
$= \text{R.H.S.}$ Showed.