❓ ୧. ବନ୍ଧନୀ ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।

(a) $\sin 80^\circ = \dots$ $[\sin 10^\circ, \sin 20^\circ, \cos 10^\circ, \cos 20^\circ]$ ।

(b) $\cos 65^\circ = \dots$ $[\sin 25^\circ, \sin 35^\circ, \cos 25^\circ, \cos 35^\circ]$ ।

(c) $\sin 180^\circ = \dots$ $[1, -1, 0, \pm 1]$ ।

(d) $\cos 90^\circ = \dots$ $[1, -1, 0, \pm 1]$ ।

(e) $\cos 110^\circ + \sin 20^\circ = \dots$ $[2 \cos 110^\circ, 2 \sin 20^\circ, 0, 1]$ ।

(f) $\sin 75^\circ - \cos 15^\circ = \dots$ $[\frac{\sqrt{3}}{2}, \frac{1}{2}, 0, 1]$ ।

(g) $\sin 0^\circ = \dots$ $[\cos 0^\circ, \sin 90^\circ, \sin 180^\circ, \cos 180^\circ]$ ।

(h) $\sin 15^\circ + \cos 105^\circ = \dots$ $[0, 1, -1, \pm 1]$ ।

(i) $\cos 121^\circ + \sin 149^\circ = \dots$ $[1, -1, 0, \pm 1]$ ।

(j) $\tan 102^\circ - \cot 168^\circ = \dots$ $[0, -1, 1, \pm 1]$ ।

✅ ଉତ୍ତର:

(a) $\cos 10^\circ$ (କାରଣ $\sin 80^\circ = \sin(90^\circ-10^\circ) = \cos 10^\circ$)
(b) $\sin 25^\circ$ (କାରଣ $\cos 65^\circ = \cos(90^\circ-25^\circ) = \sin 25^\circ$)
(c)  0 
(d)  0   
(e) 0 (କାରଣ $\cos 110^\circ = \cos(90^\circ+20^\circ) = -\sin 20^\circ$)
(f) 0 (କାରଣ $\sin 75^\circ = \sin(90^\circ-15^\circ) = \cos 15^\circ$)
(g)  $\sin 180^\circ$ (କାରଣ $\sin 0^\circ = 0$ ଏବଂ $\sin 180^\circ = 0$)   
(h) 0 (କାରଣ $\cos 105^\circ = \cos(90^\circ+15^\circ) = -\sin 15^\circ$)
(i) 0 (କାରଣ $\cos 121^\circ = -\sin 31^\circ$ ଏବଂ $\sin 149^\circ = \sin(180^\circ-31^\circ) = \sin 31^\circ$)
(j) 0 (କାରଣ $\tan 102^\circ = -\cot 12^\circ$ ଏବଂ $\cot 168^\circ = -\cot 12^\circ$)

❓ ୨. $90^\circ + \theta$ କିମ୍ବା $90^\circ - \theta$ କିମ୍ବା $180^\circ - \theta$, ର ତ୍ରିକୋଣମିତିକ ଅନୁପାତ ରୂପରେ ପ୍ରକାଶ କର 

(i) $\sin 111^\circ$ (ii) $\cos 122^\circ$ (iii) $\tan 99^\circ$ (iv) $\cot 101^\circ$ (v) $\sin 91^\circ$ (vi) $\csc 93^\circ$ (vii) $\cos 128^\circ$ (viii) $\csc 132^\circ$ (ix) $\cot 131^\cir$ 

✅ ଉତ୍ତର: (i) $\sin 111^\circ = \sin(90^\circ + 21^\circ) = \cos 21^\circ$
(ii) $\cos 122^\circ = \cos(90^\circ + 32^\circ) = -\sin 32^\circ$
(iii) $\tan 99^\circ = \tan(90^\circ + 9^\circ) = -\cot 9^\circ$
(iv) $\cot 101^\circ = \cot(90^\circ + 11^\circ) = -\tan 11^\circ$
(v) $\sin 91^\circ = \sin(90^\circ + 1^\circ) = \cos 1^\circ$
(vi) $\csc 93^\circ = \csc(90^\circ + 3^\circ) = \sec 3^\circ$
(vii) $\cos 128^\circ = \cos(90^\circ + 38^\circ) = -\sin 38^\circ$
(viii) $\csc 132^\circ = \csc(180^\circ - 48^\circ) = \csc 48^\circ$
(ix) $\cot 131^\circ = \cot(180^\circ - 49^\circ) = -\cot 49^\circ$

❓ ୩. ନିମ୍ନସ୍ତ ପଦଗୁଡ଼ିକୁ $0^\circ$ ଏବଂ $45^\circ$ କୋଣ ପରିମାଣ ମଧ୍ୟସ୍ଥ ତ୍ରିକୋଣମିତିକ ଅନୁପାତରେ ପ୍ରକାଶ କର ।

(i) $\cos 85^\circ + \cot 85^\circ$ । (ii) $\sin 75^\circ + \tan 75^\circ$ । (iii) $\cot 65^\circ + \tan 49^\circ$ ।

✅ ଉତ୍ତର: (i) $\cos 85^\circ + \cot 85^\circ = \cos(90^\circ - 5^\circ) + \cot(90^\circ - 5^\circ) = \sin 5^\circ + \tan 5^\circ$
(ii) $\sin 75^\circ + \tan 75^\circ = \sin(90^\circ - 15^\circ) + \tan(90^\circ - 15^\circ) = \cos 15^\circ + \cot 15^\circ$
(iii) $\cot 65^\circ + \tan 49^\circ = \cot(90^\circ - 25^\circ) + \tan(90^\circ - 41^\circ) = \tan 25^\circ + \cot 41^\circ$

❓ ୪. ମାନ ନିର୍ଣ୍ଣୟ କର । (i) $\frac{\sin 18^\circ}{\cos 72^\circ}$ (ii) $\frac{\tan 26^\circ}{\cot 64^\circ}$

(iii) $\frac{\sin 116^\circ}{\cos 26^\circ}$ (iv) $\frac{\csc 74^\circ}{\csc 106^\circ}$ (v) $\frac{\sin 28^\circ}{\cos 118^\circ}$ ।

✅ ଉତ୍ତର: (i) $\frac{\sin 18^\circ}{\cos 72^\circ} = \frac{\sin(90^\circ - 72^\circ)}{\cos 72^\circ} = \frac{\cos 72^\circ}{\cos 72^\circ} = 1$
(ii) $\frac{\tan 26^\circ}{\cot 64^\circ} = \frac{\tan(90^\circ - 64^\circ)}{\cot 64^\circ} = \frac{\cot 64^\circ}{\cot 64^\circ} = 1$
(iii) $\frac{\sin 116^\circ}{\cos 26^\circ} = \frac{\sin(90^\circ + 26^\circ)}{\cos 26^\circ} = \frac{\cos 26^\circ}{\cos 26^\circ} = 1$
(iv) $\frac{\csc 74^\circ}{\csc 106^\circ} = \frac{\csc 74^\circ}{\csc(180^\circ - 74^\circ)} = \frac{\csc 74^\circ}{\csc 74^\circ} = 1$
(v) $\frac{\sin 28^\circ}{\cos 118^\circ} = \frac{\sin 28^\circ}{\cos(90^\circ + 28^\circ)} = \frac{\sin 28^\circ}{-\sin 28^\circ} = -1$

❓ ୫. ସରଳ କର :- (i) $\csc 31^\circ - \sec 59^\circ$

(ii) $\sin(50^\circ + \theta) - \cos(40^\circ - \theta)$

(iii) $\frac{\cos^2 20^\circ + \cos^2 70^\circ}{\sin^2 59^\circ + \sin^2 31^\circ}$

(iv) $\tan(55^\circ - \theta) - \cot(35^\circ + \theta)$

(v) $\cos 1^\circ \cdot \cos 2^\circ \dots \cos 180^\circ$

(vi) $(\frac{\sin 27^\circ}{\cos 63^\circ})^2 + (\frac{\cos 63^\circ}{\sin 27^\circ})^2$

(vii) $\cot 112^\circ \cdot \cot 158^\circ$

(viii) $\cos^2(90^\circ + \alpha) + \cos^2(180^\circ - \alpha)$

(ix) $\sec^2(105^\circ + \alpha) - \tan^2(75^\circ - \alpha)$

(x) $\sin^2(110^\circ + \alpha) + \cos^2(70^\circ - \alpha)$ ।

✅ ଉତ୍ତର: (i) $\csc 31^\circ - \sec 59^\circ = \csc 31^\circ - \csc(90^\circ - 59^\circ) = \csc 31^\circ - \csc 31^\circ = 0$
(ii) $\sin(50^\circ + \theta) - \cos(40^\circ - \theta) = \sin(50^\circ + \theta) - \sin(90^\circ - (40^\circ - \theta)) = \sin(50^\circ + \theta) - \sin(50^\circ + \theta) = 0$
(iii) $\frac{\cos^2 20^\circ + \cos^2 70^\circ}{\sin^2 59^\circ + \sin^2 31^\circ} = \frac{\cos^2 20^\circ + \sin^2 20^\circ}{\sin^2 59^\circ + \cos^2 59^\circ} = \frac{1}{1} = 1$
(iv) $\tan(55^\circ - \theta) - \cot(35^\circ + \theta) = \tan(55^\circ - \theta) - \tan(90^\circ - (35^\circ + \theta)) = \tan(55^\circ - \theta) - \tan(55^\circ - \theta) = 0$
(v) ଏହି ଗୁଣଫଳରେ $\cos 90^\circ$ ଅଛି
ଯେହେତୁ $\cos 90^\circ = 0$, ତେଣୁ ସମୁଦାୟ ଗୁଣଫଳ $0$ ହେବ
(vi) $(\frac{\sin 27^\circ}{\cos 63^\circ})^2 + (\frac{\cos 63^\circ}{\sin 27^\circ})^2 = (1)^2 + (1)^2 = 1 + 1 = 2$
(vii) $\cot 112^\circ \cdot \cot 158^\circ = \cot(90^\circ + 22^\circ) \cdot \cot(180^\circ - 22^\circ) = (-\tan 22^\circ) \cdot (-\cot 22^\circ) = \tan 22^\circ \cdot \cot 22^\circ = 1$
(viii) $\cos^2(90^\circ + \alpha) + \cos^2(180^\circ - \alpha) = (-\sin \alpha)^2 + (-\cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha = 1$
(ix) $\sec^2(105^\circ + \alpha) - \tan^2(75^\circ - \alpha) = \csc^2(15^\circ - \alpha) - \cot^2(15^\circ - \alpha) = 1$
(x) $\sin^2(110^\circ + \alpha) + \cos^2(70^\circ - \alpha) = \cos^2(20^\circ + \alpha) + \sin^2(20^\circ + \alpha) = 1$

❓ ୬. ମାନ ନିର୍ଣ୍ଣୟ କର । (i) $\csc^2 67^\circ - \tan^2 23^\circ$ (ii) $\frac{\sin 51^\circ + \sin 156^\circ}{\cos 39^\circ + \cos 66^\circ}$ (iii) $\frac{\cos 68^\circ + \sin 131^\circ}{\sin 22^\circ + \cos 41^\circ}$ (iv) $\frac{\sin 162^\circ + \cos 153^\circ}{\cos 72^\circ - \cos 27^\circ}$ (v) $\frac{\cos 38^\circ + \sin 120^\circ}{2 \sin 52^\circ + \sqrt{3}}$ (vi) $\frac{2 \cos 67^\circ}{\sin 23^\circ} - \frac{\tan 40^\circ}{\cot 50^\circ} - \sin 90^\circ$ (vii) $\frac{\sec 61^\circ + \csc 120^\circ}{\sqrt{3} \csc 29^\circ + 2}$ ।

✅ ଉତ୍ତର: (i) $\csc^2 67^\circ - \tan^2 23^\circ = \sec^2 23^\circ - \tan^2 23^\circ = 1$
(ii) $\frac{\sin 51^\circ + \sin(90^\circ + 66^\circ)}{\cos(90^\circ - 51^\circ) + \cos 66^\circ} = \frac{\sin 51^\circ + \cos 66^\circ}{\sin 51^\circ + \cos 66^\circ} = 1$
(iii) $\frac{\cos 68^\circ + \sin(90^\circ + 41^\circ)}{\sin(90^\circ - 68^\circ) + \cos 41^\circ} = \frac{\cos 68^\circ + \cos 41^\circ}{\cos 68^\circ + \cos 41^\circ} = 1$
(iv) $\frac{\sin 162^\circ + \cos 153^\circ}{\cos 72^\circ - \cos 27^\circ} = \frac{\sin(180^\circ - 18^\circ) + \cos(180^\circ - 27^\circ)}{\sin(90^\circ - 72^\circ) - \cos 27^\circ} = \frac{\sin 18^\circ - \cos 27^\circ}{\sin 18^\circ - \cos 27^\circ} = 1$
(v) $\frac{\cos 38^\circ + \sin 120^\circ}{2 \sin 52^\circ + \sqrt{3}} = \frac{\sin 52^\circ + \sqrt{3}/2}{2(\sin 52^\circ + \sqrt{3}/2)} = \frac{1}{2}$
(vi) $\frac{2 \cos 67^\circ}{\sin 23^\circ} - \frac{\tan 40^\circ}{\cot 50^\circ} - \sin 90^\circ = \frac{2 \sin 23^\circ}{\sin 23^\circ} - \frac{\tan 40^\circ}{\tan 40^\circ} - 1 = 2 - 1 - 1 = 0$
(vii) $\frac{\sec 61^\circ + \csc 120^\circ}{\sqrt{3} \csc 29^\circ + 2} = \frac{\csc 29^\circ + 2/\sqrt{3}}{\sqrt{3}(\csc 29^\circ + 2/\sqrt{3})} = \frac{1}{\sqrt{3}}$

• $\csc^2 \theta - \cot^2 \theta = 1$

❓ ୭. ପ୍ରମାଣ କର : (i) $\cos(90^\circ-\theta) \cdot \csc(180^\circ-\theta) = 1$

(ii) $\frac{\cos 29^\circ + \sin 159^\circ}{\sin 61^\circ + \cos 69^\circ} = 1$

(iii) $\sin^2 70^\circ + \cos^2 110^\circ = 1$

(iv) $\sin^2 110^\circ + \sin^2 20^\circ = 1$

(v) $\sec^2 \theta + \csc^2(180^\circ-\theta) = \sec^2 \theta \cdot \csc^2 \theta$

(vi) $2\sin \theta \cdot \sec(90^\circ+\theta) \cdot \sin 30^\circ \cdot \tan 135^\circ = 1$

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= \cos(90^\circ-\theta) \cdot \csc(180^\circ-\theta) = \sin \theta \cdot \csc \theta = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ବାମପକ୍ଷ $= \frac{\cos 29^\circ + \sin(180^\circ-21^\circ)}{\sin(90^\circ-29^\circ) + \cos(90^\circ-21^\circ)} = \frac{\cos 29^\circ + \sin 21^\circ}{\cos 29^\circ + \sin 21^\circ} = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

(iii) ବାମପକ୍ଷ $= \sin^2 70^\circ + \cos^2(180^\circ-70^\circ) = \sin^2 70^\circ + (-\cos 70^\circ)^2 = \sin^2 70^\circ + \cos^2 70^\circ = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

(iv) ବାମପକ୍ଷ $= \sin^2(90^\circ+20^\circ) + \sin^2 20^\circ = \cos^2 20^\circ + \sin^2 20^\circ = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

(v) ବାମପକ୍ଷ $= \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cdot \cos^2 \theta} = \frac{1}{\sin^2 \theta \cdot \cos^2 \theta} = \sec^2 \theta \cdot \csc^2 \theta =$ ଦକ୍ଷିଣପକ୍ଷ

(vi) ବାମପକ୍ଷ $= 2\sin \theta \cdot (-\csc \theta) \cdot \frac{1}{2} \cdot (-1) = -2(\sin \theta \cdot \csc \theta) \cdot \frac{1}{2} \cdot (-1) = -2 \cdot 1 \cdot -\frac{1}{2} = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

❓ ୮. ପ୍ରମାଣ କର :

(i) $\cos^2 135^\circ - 2\sin^2 180^\circ + 3\cot^2 150^\circ - 4\tan^2 120^\circ = -\frac{5}{2}$

(ii) $\tan 30^\circ \cdot \tan 135^\circ \cdot \tan 150^\circ \cdot \tan 45^\circ = 1$

(iii) $\frac{\sec^2 180^\circ + \tan 150^\circ}{\csc^2 90^\circ + \cot 120^\circ} = 1$

(iv) $\sin^2 135^\circ + \cos^2 120^\circ - \sin^2 120^\circ + \tan^2 150^\circ = \frac{1}{3}$

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= (-\frac{1}{\sqrt{2}})^2 - 2(0)^2 + 3(-\sqrt{3})^2 - 4(-\sqrt{3})^2 = \frac{1}{2} - 0 + 9 - 12 = \frac{1}{2} - 3 = -\frac{5}{2} =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ବାମପକ୍ଷ $= \frac{1}{\sqrt{3}} \cdot (-1) \cdot (-\frac{1}{\sqrt{3}}) \cdot 1 = \frac{1}{3}$ (ସୂଚନା: ପୁସ୍ତକରେ ଦିଆଯାଇଥିବା ଉତ୍ତର 1 ରେ ସାମାନ୍ୟ ତ୍ରୁଟି ଥାଇପାରେ)

(iii) ବାମପକ୍ଷ $= \frac{(-1)^2 + (-1/\sqrt{3})}{1^2 + (-1/\sqrt{3})} = \frac{1 - 1/\sqrt{3}}{1 - 1/\sqrt{3}} = 1 =$ ଦକ୍ଷିଣପକ୍ଷ

(iv) ବାମପକ୍ଷ $= (\frac{1}{\sqrt{2}})^2 + (-\frac{1}{2})^2 - (\frac{\sqrt{3}}{2})^2 + (-\frac{1}{\sqrt{3}})^2 = \frac{1}{2} + \frac{1}{4} - \frac{3}{4} + \frac{1}{3} = \frac{2+1-3}{4} + \frac{1}{3} = 0 + \frac{1}{3} = \frac{1}{3} =$ ଦକ୍ଷିଣପକ୍ଷ 

❓ 9. ମୂଲ୍ୟ ନିରୂପଣ କର :

(i) $\tan 10^\circ \cdot \tan 20^\circ \cdot \tan 30^\circ \dots \tan 70^\circ \cdot \tan 80^\circ$

(ii) $\cot 12^\circ \cdot \cot 38^\circ \cdot \cot 52^\circ \cdot \cot 60^\circ \cdot \cot 78^\circ$

(iii) $\tan 5^\circ \cdot \tan 15^\circ \cdot \tan 45^\circ \cdot \tan 75^\circ \cdot \tan 85^\circ$

✅ ଉତ୍ତର: (i) $(\tan 10^\circ \cdot \tan 80^\circ) \cdot (\tan 20^\circ \cdot \tan 70^\circ) \cdot (\tan 30^\circ \cdot \tan 60^\circ) \cdot \tan 40^\circ \cdot \tan 50^\circ = (\tan 10^\circ \cdot \cot 10^\circ) \cdot (\tan 20^\circ \cdot \cot 20^\circ) \cdot (\tan 30^\circ \cdot \cot 30^\circ) \cdot (\tan 40^\circ \cdot \cot 40^\circ) = 1 \cdot 1 \cdot 1 \cdot 1 = 1$

(ii) $(\cot 12^\circ \cdot \cot 78^\circ) \cdot (\cot 38^\circ \cdot \cot 52^\circ) \cdot \cot 60^\circ = (\cot 12^\circ \cdot \tan 12^\circ) \cdot (\cot 38^\circ \cdot \tan 38^\circ) \cdot \frac{1}{\sqrt{3}} = 1 \cdot 1 \cdot \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$

(iii) $(\tan 5^\circ \cdot \tan 85^\circ) \cdot (\tan 15^\circ \cdot \tan 75^\circ) \cdot \tan 45^\circ = (\tan 5^\circ \cdot \cot 5^\circ) \cdot (\tan 15^\circ \cdot \cot 15^\circ) \cdot 1 = 1 \cdot 1 \cdot 1 = 1$

❓ ୧୦. ପ୍ରମାଣ କର :

(i) $\sin 120^{\circ} + \tan 150^{\circ} \cdot \cos 135^{\circ} = \frac{3+\sqrt{2}}{2\sqrt{3}}$

(ii) $\frac{\sec^2 180^{\circ} + \tan 150^{\circ}}{\csc^2 90^{\circ} + \cot 120^{\circ}} = 2-\sqrt{3}$

(iii) $\frac{\sec^2 180^{\circ} + \tan 45^{\circ}}{\csc^2 90^{\circ} - \cot 120^{\circ}} = 3-\sqrt{3}$ ।

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= \sin 120^{\circ} + \tan 150^{\circ} \cdot \cos 135^{\circ}$
$= \sin(180^{\circ}-60^{\circ}) + \tan(180^{\circ}-30^{\circ}) \cdot \cos(180^{\circ}-45^{\circ})$
$= \sin 60^{\circ} + (-\tan 30^{\circ}) \cdot (-\cos 45^{\circ})$
$= \frac{\sqrt{3}}{2} + \left(-\frac{1}{\sqrt{3}}\right) \cdot \left(-\frac{1}{\sqrt{2}}\right) = \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{6}}$
$= \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2} \cdot \sqrt{3}} = \frac{3 + \sqrt{2}}{2\sqrt{3}}$
(ii) ବାମପକ୍ଷ $= \frac{(-1)^2 + (-1/\sqrt{3})}{1^2 + (-1/\sqrt{3})} = 1$ (ସୂଚନା: ପୁସ୍ତକରେ ଦକ୍ଷିଣପକ୍ଷ $2-\sqrt{3}$ ପାଇବା ପାଇଁ ହରରେ ଚିହ୍ନ '-' ହେବା ଆବଶ୍ୟକ)
(iii) ବାମପକ୍ଷ $= \frac{(-1)^2 + 1}{1^2 - (-1/\sqrt{3})} = \frac{2}{1 + \frac{1}{\sqrt{3}}} = \frac{2\sqrt{3}}{\sqrt{3}+1}$
$= \frac{2\sqrt{3}(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{2(3-\sqrt{3})}{2} = 3-\sqrt{3}$ (ପ୍ରମାଣିତ)

❓ ୧୧. ସରଳ କର : (i) $\sin(180^{\circ}-\theta) \cdot \cos(90^{\circ}+\theta) + \sin(90^{\circ}+\theta) \cdot \cos(180^{\circ}-\theta)$

(ii) $\frac{\cos(90^{\circ}-A) \cdot \sec(180^{\circ}-A) \cdot \sin(180^{\circ}-A)}{\sin(90^{\circ}+A) \cdot \tan(90^{\circ}+A) \cdot \csc(90^{\circ}+A)}$ ।

✅ ଉତ୍ତର: (i) $\sin \theta \cdot (-\sin \theta) + \cos \theta \cdot (-\cos \theta)$
$= -(\sin^2 \theta + \cos^2 \theta) = -1$
(ii) ଲବ $= \sin A \cdot (-\sec A) \cdot \sin A = -\sin^2 A \cdot \sec A = -\frac{\sin^2 A}{\cos A}$
ହର $= \cos A \cdot (-\cot A) \cdot \sec A = -\cot A$
ସମ୍ପୂର୍ଣ୍ଣ ରାଶି $= \frac{-\sin^2 A / \cos A}{-\cos A / \sin A} = \frac{\sin^3 A}{\cos^2 A}$

❓ ୧୨. $\Delta ABC$ ରେ $m\angle B=90^{\circ}$ ହେଲେ ପ୍ରମାଣ କର ଯେ, $\sin^2 A + \sin^2 C = 1$ ।

✅ ଉତ୍ତର: $\Delta ABC$ ରେ $A + B + C = 180^{\circ}$
$B = 90^{\circ}$ ହେତୁ $A + C = 90^{\circ} \Rightarrow C = 90^{\circ} - A$
ବାମପକ୍ଷ $= \sin^2 A + \sin^2(90^{\circ} - A) = \sin^2 A + \cos^2 A = 1$ (ପ୍ରମାଣିତ)

❓ ୧୩. $\Delta ABC$ ରେ ପ୍ରମାଣ କର ଯେ, $\cos(A+B) + \sin C = \sin(A+B) - \cos C$ ।

✅ ଉତ୍ତର: ଆମେ ଜାଣୁ $A + B + C = 180^{\circ} \Rightarrow A + B = 180^{\circ} - C$
ବାମପକ୍ଷ $= \cos(180^{\circ} - C) + \sin C = -\cos C + \sin C$
ଦକ୍ଷିଣପକ୍ଷ $= \sin(180^{\circ} - C) - \cos C = \sin C - \cos C$
ତେଣୁ ବାମପକ୍ଷ $=$ ଦକ୍ଷିଣପକ୍ଷ (ପ୍ରମାଣିତ)

❓ ୧୪. A ଓ B ଦୁଇଟି ପରସ୍ପର ଅନୁପୁରକ କୋଣ ହେଲେ $\sin A \cdot \cos B + \cos A \cdot \sin B$ ର ମାନ ନିର୍ଣ୍ଣୟ କର ।

✅ ଉତ୍ତର: ଦତ୍ତ ଅଛି $A + B = 90^{\circ} \Rightarrow B = 90^{\circ} - A$
$\sin A \cdot \cos(90^{\circ}-A) + \cos A \cdot \sin(90^{\circ}-A)$
$= \sin A \cdot \sin A + \cos A \cdot \cos A = \sin^2 A + \cos^2 A = 1$

❓ ୧୫. ABCD ଏକ ବୃତ୍ତାନ୍ତର୍ଲିଖତ ଚତୁର୍ଭୁଜ ହେଲେ $\tan A + \tan C$ ର ମାନ ନିର୍ଣ୍ଣୟ କର ।

✅ ଉତ୍ତର: ବୃତ୍ତାନ୍ତର୍ଲିଖତ ଚତୁର୍ଭୁଜରେ ବିପରୀତ କୋଣମାନଙ୍କର ସମଷ୍ଟି $180^{\circ}$
ତେଣୁ $A + C = 180^{\circ} \Rightarrow C = 180^{\circ} - A$
$\tan A + \tan(180^{\circ} - A) = \tan A - \tan A = 0$