❓ ୧. ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର । (i) $\sin(A-B) = \frac{\sin A}{.......} - \frac{\cos A}{.......}$ 

(ii) $\cos(\theta+\alpha) + \cos(\alpha-\theta) = ...........$ 

(iii) $\cos(60^\circ-A) + ............. = \cos A$ 

(iv) $\sin(30^\circ+A) + \sin(30^\circ-A) = ......$ 

(v) $2 \sin A \sin B = ........ - \cos(A+B)$ 

(vi) $\tan(45^\circ+\theta) \cdot \tan(45^\circ-\theta) = ................$ 

✅ ଉତ୍ତର: (i) $\csc B$ ଏବଂ $\sec B$

ଆମେ ଜାଣୁ $\sin(A-B) = \sin A \cos B - \cos A \sin B$

ଏହାକୁ $\frac{\sin A}{\sec B} - \frac{\cos A}{\csc B}$ ଭାବେ ଲେଖାଯାଇପାରିବ

(ii) $2 \cos \alpha \cos \theta$

ସୂତ୍ର ଅନୁଯାୟୀ $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$

(iii) $\cos(60^\circ+A)$

କାରଣ $\cos(60^\circ-A) + \cos(60^\circ+A) = 2 \cos 60^\circ \cos A = 2 \cdot \frac{1}{2} \cdot \cos A = \cos A$

(iv) $\cos A$

କାରଣ $\sin(30^\circ+A) + \sin(30^\circ-A) = 2 \sin 30^\circ \cos A = 2 \cdot \frac{1}{2} \cdot \cos A = \cos A$

(v) $\cos(A-B)$

ସୂତ୍ର ଅନୁଯାୟୀ $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$

(vi) $1$

କାରଣ $\frac{1+\tan \theta}{1-\tan \theta} \cdot \frac{1-\tan \theta}{1+\tan \theta} = 1$

❓ ୨. ପ୍ରମାଣ କର :

(i) $\frac{\sin(A-B)}{\cos A \cos B} = \tan A - \tan B$ 

(ii) $\frac{\cos(A+B)}{\cos A \cos B} = 1 - \tan A \tan B$ 

(iii) $\frac{\cos(A-B)}{\cos A \sin B} = \cot B + \tan A$ 

(iv) $\frac{\sin \alpha}{\sin \beta} - \frac{\cos \alpha}{\cos \beta} = \frac{\sin(\alpha-\beta)}{\sin \beta \cos \beta}$ 

(v) $\frac{\cos \alpha}{\sin \beta} - \frac{\sin \alpha}{\cos \beta} = \frac{\cos(\alpha+\beta)}{\sin \beta \cos \beta}$ 

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= \frac{\sin(A-B)}{\cos A \cos B} = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} = \frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B} = \tan A - \tan B =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ବାମପକ୍ଷ $= \frac{\cos(A+B)}{\cos A \cos B} = \frac{\cos A \cos B - \sin A \sin B}{\cos A \cos B} = \frac{\cos A \cos B}{\cos A \cos B} - \frac{\sin A \sin B}{\cos A \cos B} = 1 - \tan A \tan B =$ ଦକ୍ଷିଣପକ୍ଷ

(iii) ବାମପକ୍ଷ $= \frac{\cos(A-B)}{\cos A \sin B} = \frac{\cos A \cos B + \sin A \sin B}{\cos A \sin B} = \frac{\cos A \cos B}{\cos A \sin B} + \frac{\sin A \sin B}{\cos A \sin B} = \cot B + \tan A =$ ଦକ୍ଷିଣପକ୍ଷ

(iv) ବାମପକ୍ଷ $= \frac{\sin \alpha}{\sin \beta} - \frac{\cos \alpha}{\cos \beta} = \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\sin \beta \cos \beta} = \frac{\sin(\alpha-\beta)}{\sin \beta \cos \beta} =$ ଦକ୍ଷିଣପକ୍ଷ

(v) ବାମପକ୍ଷ $= \frac{\cos \alpha}{\sin \beta} - \frac{\sin \alpha}{\cos \beta} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \beta \cos \beta} = \frac{\cos(\alpha+\beta)}{\sin \beta \cos \beta} =$ ଦକ୍ଷିଣପକ୍ଷ

❓ ୩. ପ୍ରମାଣ କର : 

(i) $\cos(A+45^\circ) = \frac{1}{\sqrt{2}}(\cos A - \sin A)$ 

(ii) $\sin(45^\circ-\theta) = -\frac{1}{\sqrt{2}}(\sin \theta - \cos \theta)$ 

(iii) $\tan(45^\circ+\theta) = \frac{1+\tan \theta}{1-\tan \theta}$ 

(iv) $\cot(45^\circ-\theta) = \frac{\cot \theta + 1}{\cot \theta - 1}$ 

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= \cos(A+45^\circ) = \cos A \cos 45^\circ - \sin A \sin 45^\circ = \cos A \cdot \frac{1}{\sqrt{2}} - \sin A \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\cos A - \sin A) =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ବାମପକ୍ଷ $= \sin(45^\circ-\theta) = \sin 45^\circ \cos \theta - \cos 45^\circ \sin \theta = \frac{1}{\sqrt{2}}\cos \theta - \frac{1}{\sqrt{2}}\sin \theta = \frac{1}{\sqrt{2}}(\cos \theta - \sin \theta) = -\frac{1}{\sqrt{2}}(\sin \theta - \cos \theta) =$ ଦକ୍ଷିଣପକ୍ଷ

(iii) ବାମପକ୍ଷ $= \tan(45^\circ+\theta) = \frac{\tan 45^\circ + \tan \theta}{1 - \tan 45^\circ \tan \theta} = \frac{1+\tan \theta}{1 - \tan \theta} =$ ଦକ୍ଷିଣପକ୍ଷ

(iv) ବାମପକ୍ଷ $= \cot(45^\circ-\theta) = \frac{\cot 45^\circ \cot \theta + 1}{\cot \theta - \cot 45^\circ} = \frac{1 \cdot \cot \theta + 1}{\cot \theta - 1} = \frac{\cot \theta + 1}{\cot \theta - 1} =$ ଦକ୍ଷିଣପକ୍ଷ

❓ ୪. ପ୍ରମାଣ କର :

(i) $\cos(45^\circ-A)\cos(45^\circ-B) - \sin(45^\circ-A)\sin(45^\circ-B) = \sin(A+B)$ 

(ii) $\sin(40^\circ+A)\cos(20^\circ-A) + \cos(40^\circ+A)\sin(20^\circ-A) = \frac{\sqrt{3}}{2}$ 

(iii) $\cos(65^\circ+\theta)\cos(35^\circ+\theta) + \sin(65^\circ+\theta)\sin(35^\circ+\theta) = \frac{\sqrt{3}}{2}$ 

(iv) $\cos n\theta \cos \theta + \sin n\theta \sin \theta = \cos(n-1)\theta$ 

(v) $\tan(60^\circ-A) = \frac{\sqrt{3} \cos A - \sin A}{\cos A + \sqrt{3} \sin A}$ ।

✅ ଉତ୍ତର: (i) ମନେକର $X = 45^\circ-A$ ଏବଂ $Y = 45^\circ-B$

ବାମପକ୍ଷ $= \cos X \cos Y - \sin X \sin Y = \cos(X+Y) = \cos(45^\circ-A + 45^\circ-B) = \cos(90^\circ - (A+B)) = \sin(A+B) =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ମନେକର $X = 40^\circ+A$ ଏବଂ $Y = 20^\circ-A$

ବାମପକ୍ଷ $= \sin X \cos Y + \cos X \sin Y = \sin(X+Y) = \sin(40^\circ+A + 20^\circ-A) = \sin 60^\circ = \frac{\sqrt{3}}{2} =$ ଦକ୍ଷିଣପକ୍ଷ

(iii) ମନେକର $X = 65^\circ+\theta$ ଏବଂ $Y = 35^\circ+\theta$

ବାମପକ୍ଷ $= \cos X \cos Y + \sin X \sin Y = \cos(X-Y) = \cos(65^\circ+\theta - (35^\circ+\theta)) = \cos 30^\circ = \frac{\sqrt{3}}{2} =$ ଦକ୍ଷିଣପକ୍ଷ

(iv) ବାମପକ୍ଷ $= \cos n\theta \cos \theta + \sin n\theta \sin \theta = \cos(n\theta - \theta) = \cos(n-1)\theta =$ ଦକ୍ଷିଣପକ୍ଷ

(v) ବାମପକ୍ଷ $= \tan(60^\circ-A) = \frac{\tan 60^\circ - \tan A}{1 + \tan 60^\circ \tan A} = \frac{\sqrt{3} - \frac{\sin A}{\cos A}}{1 + \sqrt{3} \cdot \frac{\sin A}{\cos A}} = \frac{\frac{\sqrt{3} \cos A - \sin A}{\cos A}}{\frac{\cos A + \sqrt{3} \sin A}{\cos A}} = \frac{\sqrt{3} \cos A - \sin A}{\cos A + \sqrt{3} \sin A} =$ ଦକ୍ଷିଣପକ୍ଷ

❓ ୫. ପ୍ରମାଣ କର :

(i) $\tan 62^\circ = \frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ}$ ।

(ii) $\frac{\cos 25^\circ + \sin 25^\circ}{\cos 25^\circ - \sin 25^\circ} = \tan 70^\circ$ ।

(iii) $\tan 7A - \tan 4A - \tan 3A = \tan 7A \cdot \tan 4A \cdot \tan 3A$ ।

(iv) $\tan(x+y) - \tan x - \tan y = \tan(x+y) \cdot \tan x \cdot \tan y$ ।

(v) $(1 + \tan 15^\circ)(1 + \tan 30^\circ) = 2$ ।

(vi) $(\cot 10^\circ - 1)(\cot 35^\circ - 1) = 2$ ।

(vii) $\frac{1}{\cot A + \tan B} - \frac{1}{\tan A + \cot B} = \tan(A - B)$ ।

(viii) $\sqrt{3} + \cot 50^\circ + \tan 80^\circ = \sqrt{3} \cot 50^\circ \cdot \tan 80^\circ$ ।

✅ ଉତ୍ତର: (i) ବାମପକ୍ଷ $= \tan 62^\circ = \tan(45^\circ + 17^\circ)$

ଆମେ ଜାଣୁ $\tan(45^\circ + \theta) = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta}$

ତେଣୁ $\tan(45^\circ + 17^\circ) = \frac{\cos 17^\circ + \sin 17^\circ}{\cos 17^\circ - \sin 17^\circ} =$ ଦକ୍ଷିଣପକ୍ଷ

(ii) ବାମପକ୍ଷ $= \frac{\cos 25^\circ + \sin 25^\circ}{\cos 25^\circ - \sin 25^\circ}$

ସୂତ୍ର ଅନୁଯାୟୀ ଏହା $\tan(45^\circ + 25^\circ)$ ସହ ସମାନ

ତେଣୁ $\tan(45^\circ + 25^\circ) = \tan 70^\circ =$ ଦକ୍ଷିଣପକ୍ଷ

(iii) ଆମେ ଜାଣୁ $7A = 4A + 3A$

ଉଭୟ ପାର୍ଶ୍ୱରେ $\tan$ ନେଲେ, $\tan 7A = \tan(4A + 3A) = \frac{\tan 4A + \tan 3A}{1 - \tan 4A \tan 3A}$

ବ୍ରଜ୍ର ଗୁଣନ କଲେ, $\tan 7A (1 - \tan 4A \tan 3A) = \tan 4A + \tan 3A$

$\tan 7A - \tan 7A \tan 4A \tan 3A = \tan 4A + \tan 3A$

$\tan 7A - \tan 4A - \tan 3A = \tan 7A \cdot \tan 4A \cdot \tan 3A$ (ପ୍ରମାଣିତ)

(iv) ଏହା ପ୍ରଶ୍ନ (iii) ପରି ଅଟେ

ଏଠାରେ $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$ ସୂତ୍ର ପ୍ରୟୋଗ କରି ସମାଧାନ କରାଯାଇପାରିବ

(v) ଯଦି $A+B = 45^\circ$, ତେବେ $(1 + \tan A)(1 + \tan B) = 2$

ଏଠାରେ $15^\circ + 30^\circ = 45^\circ$

ତେଣୁ $(1 + \tan 15^\circ)(1 + \tan 30^\circ) = 2$ (ପ୍ରମାଣିତ)

(vi) ଯଦି $A+B = 45^\circ$, ତେବେ $(\cot A - 1)(\cot B - 1) = 2$

ଏଠାରେ $10^\circ + 35^\circ = 45^\circ$

ତେଣୁ $(\cot 10^\circ - 1)(\cot 35^\circ - 1) = 2$ (ପ୍ରମାଣିତ)

(vii) ବାମପକ୍ଷ $= \frac{1}{\frac{\cos A}{\sin A} + \frac{\sin B}{\cos B}} - \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos B}{\sin B}}$

$= \frac{\sin A \cos B}{\cos A \cos B + \sin A \sin B} - \frac{\cos A \sin B}{\sin A \sin B + \cos A \cos B}$

$= \frac{\sin A \cos B - \cos A \sin B}{\cos(A - B)} = \frac{\sin(A - B)}{\cos(A - B)} = \tan(A - B)$ (ପ୍ରମାଣିତ)

(viii) $\tan 80^\circ = \tan(30^\circ + 50^\circ) = \frac{\tan 30^\circ + \tan 50^\circ}{1 - \tan 30^\circ \tan 50^\circ} = \frac{\frac{1}{\sqrt{3}} + \tan 50^\circ}{1 - \frac{1}{\sqrt{3}} \tan 50^\circ} = \frac{1 + \sqrt{3} \tan 50^\circ}{\sqrt{3} - \tan 50^\circ}$

$\sqrt{3} \tan 80^\circ - \tan 80^\circ \tan 50^\circ = 1 + \sqrt{3} \tan 50^\circ$

$\sqrt{3} \tan 80^\circ \cot 50^\circ - \tan 80^\circ = \cot 50^\circ + \sqrt{3}$

$\sqrt{3} \tan 80^\circ \cot 50^\circ = \sqrt{3} + \cot 50^\circ + \tan 80^\circ$ (ପ୍ରମାଣିତ)

❓ ୬. $\cos 75^\circ$ ଓ $\sin 15^\circ$ ର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।

✅ ଉତ୍ତର: $\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ$

$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$

$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

❓ ୭. (i) $\cos \alpha = \frac{8}{17}$ ଓ $\sin \beta = \frac{5}{13}$ ହେଲେ $\sin(\alpha - \beta)$ ର ମାନ ନିର୍ଣ୍ଣୟ କର ।

(ii) $\tan A = \frac{1}{2}$, $\cot B = 3$ ହେଲେ $A+B$ ର ମାନ ନିର୍ଣ୍ଣୟ କର ।

(iii) $\tan \beta = \frac{1 - \tan \alpha}{1 + \tan \alpha}$ ହେଲେ, $\tan(\alpha + \beta)$ ର ମାନ ନିର୍ଣ୍ଣୟ କର ।

✅ ଉତ୍ତର: (i) $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (\frac{8}{17})^2} = \frac{15}{17}$

$\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$

$\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = (\frac{15}{17})(\frac{12}{13}) - (\frac{8}{17})(\frac{5}{13}) = \frac{180 - 40}{221} = \frac{140}{221}$

(ii) $\tan A = \frac{1}{2}$ ଏବଂ $\tan B = \frac{1}{\cot B} = \frac{1}{3}$

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - (\frac{1}{2})(\frac{1}{3})} = \frac{5/6}{5/6} = 1$

$\tan(A+B) = 1 \Rightarrow A+B = 45^\circ$

(iii) ଦତ୍ତ ଅଛି $\tan \beta = \frac{1 - \tan \alpha}{1 + \tan \alpha}$

ଆମେ ଜାଣୁ $\frac{1 - \tan \alpha}{1 + \tan \alpha} = \tan(45^\circ - \alpha)$

ତେଣୁ $\tan \beta = \tan(45^\circ - \alpha) \Rightarrow \beta = 45^\circ - \alpha \Rightarrow \alpha + \beta = 45^\circ$

ଅତଏବ $\tan(\alpha + \beta) = \tan 45^\circ = 1$

❓ ୮. $A+B+C=90^{\circ}$ ହେଲେ ପ୍ରମାଣ କର ଯେ

(i) $cot~A+cot~B+cot~C=cot~A \cdot cot~B \cdot cot~C$

(ii) $tan~A \cdot tan~B+tan~B \cdot tan~C+tan~C \cdot tan~A=1$ 

✅ ଦତ୍ତ ଅଛି $A+B+C=90^{\circ} \Rightarrow A+B = 90^{\circ} - C$
ସୂତ୍ର: $cot(A+B) = \frac{cot~A \cdot cot~B - 1}{cot~B + cot~A}$
(i) ଉଭୟ ପାର୍ଶ୍ଵରେ $cot$ ଅନୁପାତ ନେଲେ $cot(A+B) = cot(90^{\circ}-C) \Rightarrow \frac{cot~A \cdot cot~B - 1}{cot~B + cot~A} = tan~C = \frac{1}{cot~C}$
ବଜ୍ର ଗୁଣନ କଲେ $cot~C(cot~A \cdot cot~B - 1) = cot~A + cot~B \Rightarrow cot~A \cdot cot~B \cdot cot~C - cot~C = cot~A + cot~B \Rightarrow cot~A + cot~B + cot~C = cot~A \cdot cot~B \cdot cot~C$ (ପ୍ରମାଣିତ)
(ii) ଉଭୟ ପାର୍ଶ୍ଵରେ $tan$ ଅନୁପାତ ନେଲେ $tan(A+B) = tan(90^{\circ}-C) \Rightarrow \frac{tan~A + tan~B}{1 - tan~A \cdot tan~B} = cot~C = \frac{1}{tan~C}$
ବଜ୍ର ଗୁଣନ କଲେ $tan~C(tan~A + tan~B) = 1 - tan~A \cdot tan~B \Rightarrow tan~A \cdot tan~C + tan~B \cdot tan~C = 1 - tan~A \cdot tan~B \Rightarrow tan~A \cdot tan~B + tan~B \cdot tan~C + tan~C \cdot tan~A = 1$ (ପ୍ରମାଣିତ)

❓ ୯. (i) $A+B+C=180^{\circ}$ ଏବଂ $sin~C=1$ ହେଲେ ପ୍ରମାଣ କର ଯେ $tan~A \cdot tan~B=1$

(ii) $A+B+C=180^{\circ}$ ହେଲେ ପ୍ରମାଣ କର ଯେ $cot~A \cdot cot~B+cot~B \cdot cot~C+cot~C \cdot cot~A=1$

(iii) $A+B+C=180^{\circ}$ ଏବଂ $cos~A=cos~B \cdot cos~C$ ହେଲେ ପ୍ରମାଣ କର ଯେ (a) $tan~A=tan~B+tan~C$ (b) $tan~B \cdot tan~C=2$ ।

✅ (i) $sin~C=1 \Rightarrow C=90^{\circ}$
ଏଣୁ $A+B+90^{\circ}=180^{\circ} \Rightarrow A+B=90^{\circ}$
$tan~A = tan(90^{\circ}-B) = cot~B \Rightarrow tan~A \cdot tan~B = 1$ (ପ୍ରମାଣିତ)
(ii) $A+B=180^{\circ}-C \Rightarrow cot(A+B) = cot(180^{\circ}-C) \Rightarrow \frac{cot~A \cdot cot~B - 1}{cot~B + cot~A} = -cot~C$
ବଜ୍ର ଗୁଣନ କଲେ $cot~A \cdot cot~B - 1 = -cot~C(cot~B + cot~A) = -cot~B \cdot cot~C - cot~A \cdot cot~C$
$cot~A \cdot cot~B + cot~B \cdot cot~C + cot~C \cdot cot~A = 1$ (ପ୍ରମାଣିତ)
(iii-a) $cos~A = cos(180^{\circ}-(B+C)) = -cos(B+C)$
ଦତ୍ତ ଅଛି $cos~A = cos~B \cdot cos~C \Rightarrow cos~B \cdot cos~C = -(cos~B \cdot cos~C - sin~B \cdot sin~C) \Rightarrow 2~cos~B \cdot cos~C = sin~B \cdot sin~C$
$tan~B \cdot tan~C = 2$
ପୁନଶ୍ଚ $tan~A = tan(180^{\circ}-(B+C)) = -tan(B+C) = -\frac{tan~B + tan~C}{1 - tan~B \cdot tan~C} = -\frac{tan~B + tan~C}{1 - 2} = tan~B + tan~C$ (ପ୍ରମାଣିତ)
(iii-b) ପୂର୍ବ ପର୍ଯ୍ୟାୟରୁ ପ୍ରମାଣିତ ଯେ $tan~B \cdot tan~C = 2$

❓ ୧୦. ଦର୍ଶାଅ ଯେ

(i) $sin(A+B) \cdot sin(A-B)=sin^2 A-sin^2 B$

$$(ii) $cos(A+B) \cdot cos(A-B)=cos^2 A-sin^2 B$ ।

✅ ସୂତ୍ର: $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
(i) ବାମପକ୍ଷ $= (sin~A \cdot cos~B + cos~A \cdot sin~B)(sin~A \cdot cos~B - cos~A \cdot sin~B) = (sin~A \cdot cos~B)^2 - (cos~A \cdot sin~B)^2$
$= sin^2 A \cdot cos^2 B - cos^2 A \cdot sin^2 B = sin^2 A(1-sin^2 B) - (1-sin^2 A)sin^2 B = sin^2 A - sin^2 B$ (ପ୍ରମାଣିତ)
(ii) ବାମପକ୍ଷ $= (cos~A \cdot cos~B - sin~A \cdot sin~B)(cos~A \cdot cos~B + sin~A \cdot sin~B) = (cos~A \cdot cos~B)^2 - (sin~A \cdot sin~B)^2$
$= cos^2 A \cdot cos^2 B - sin^2 A \cdot sin^2 B = cos^2 A(1-sin^2 B) - (1-cos^2 A)sin^2 B = cos^2 A - sin^2 B$ (ପ୍ରମାଣିତ)

❓ ୧୧. ପ୍ରମାଣ କର :

(i) $sin~50^{\circ}+sin~40^{\circ}=\sqrt{2}sin~85^{\circ}$

(ii) $cos~50^{\circ}+cos~40^{\circ}=\sqrt{2}cos~5^{\circ}$

(iii) $sin~50^{\circ}-sin~70^{\circ}+sin~10^{\circ}=0$ ।

✅ ସୂତ୍ର: $sin(A+B)+sin(A-B) = 2~sin~A \cdot cos~B$
(i) ବାମପକ୍ଷ $= sin(45^{\circ}+5^{\circ}) + sin(45^{\circ}-5^{\circ}) = 2~sin~45^{\circ} \cdot cos~5^{\circ} = 2 \cdot \frac{1}{\sqrt{2}} \cdot cos~5^{\circ} = \sqrt{2}cos~5^{\circ} = \sqrt{2}sin(90^{\circ}-5^{\circ}) = \sqrt{2}sin~85^{\circ}$ (ପ୍ରମାଣିତ)
(ii) ସୂତ୍ର: $cos(A+B)+cos(A-B) = 2~cos~A \cdot cos~B$
ବାମପକ୍ଷ $= cos(45^{\circ}+5^{\circ}) + cos(45^{\circ}-5^{\circ}) = 2~cos~45^{\circ} \cdot cos~5^{\circ} = 2 \cdot \frac{1}{\sqrt{2}} \cdot cos~5^{\circ} = \sqrt{2}cos~5^{\circ}$ (ପ୍ରମାଣିତ)
(iii) ବାମପକ୍ଷ $= sin(30^{\circ}+20^{\circ}) + sin(30^{\circ}-20^{\circ}) - sin~70^{\circ} = 2~sin~30^{\circ} \cdot cos~20^{\circ} - sin~70^{\circ} = 2 \cdot \frac{1}{2} \cdot cos~20^{\circ} - sin~70^{\circ}$
$= cos~20^{\circ} - sin~70^{\circ} = sin(90^{\circ}-20^{\circ}) - sin~70^{\circ} = sin~70^{\circ} - sin~70^{\circ} = 0$ (ପ୍ରମାଣିତ)

❓ ୧୨. ନିମ୍ନଲିଖିତ କ୍ଷେତ୍ରରେ $A$ ଓ $B$ ର ମାନ ନିର୍ଣ୍ଣୟ କର :

(i) $sin(A+B)=\frac{1}{\sqrt{2}}, cos(A-B)=\frac{1}{\sqrt{2}}$

(ii) $cos(A+B)=-\frac{1}{2}, sin(A-B)=\frac{1}{2}$

(iii) $tan(A-B)=\frac{1}{\sqrt{3}}=cot(A+B)$

(iv) $tan(A+B)=-1, cosec(A-B)=\sqrt{2}$ ।

✅ (i) $sin(A+B) = sin~45^{\circ} \Rightarrow A+B = 45^{\circ}$ ଏବଂ $cos(A-B) = cos~45^{\circ} \Rightarrow A-B = 45^{\circ}$
ସମାଧାନ କଲେ: $A = 45^{\circ}, B = 0^{\circ}$
(ii) $cos(A+B) = cos~120^{\circ} \Rightarrow A+B = 120^{\circ}$ ଏବଂ $sin(A-B) = sin~30^{\circ} \Rightarrow A-B = 30^{\circ}$
ସମାଧାନ କଲେ: $A = 75^{\circ}, B = 45^{\circ}$
(iii) $tan(A-B) = tan~30^{\circ} \Rightarrow A-B = 30^{\circ}$ ଏବଂ $cot(A+B) = cot~60^{\circ} \Rightarrow A+B = 60^{\circ}$
ସମାଧାନ କଲେ: $A = 45^{\circ}, B = 15^{\circ}$
(iv) $tan(A+B) = tan~135^{\circ} \Rightarrow A+B = 135^{\circ}$ ଏବଂ $cosec(A-B) = cosec~45^{\circ} \Rightarrow A-B = 45^{\circ}$
ସମାଧାନ କଲେ: $A = 90^{\circ}, B = 45^{\circ}$