ক) AI-তৈরি নমুনা উত্তর
বামপক্ষ= 3 sin 20 ∘ − 1 cos 20 ∘ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = s i n 2 0 ∘ 3 − c o s 2 0 ∘ 1 = 3 cos 20 ∘ − sin 20 ∘ sin 20 ∘ cos 20 ∘ = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} = s i n 2 0 ∘ c o s 2 0 ∘ 3 c o s 2 0 ∘ − s i n 2 0 ∘ = 2 ( 3 2 cos 20 ∘ − 1 2 sin 20 ∘ ) 1 2 ( 2 sin 20 ∘ cos 20 ∘ ) = \frac{2 \left(\frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ\right)}{\frac{1}{2} (2\sin 20^\circ \cos 20^\circ)} = 2 1 ( 2 s i n 2 0 ∘ c o s 2 0 ∘ ) 2 ( 2 3 c o s 2 0 ∘ − 2 1 s i n 2 0 ∘ ) = 4 ( sin 60 ∘ cos 20 ∘ − cos 60 ∘ sin 20 ∘ ) sin ( 2 × 20 ∘ ) = \frac{4(\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\sin(2 \times 20^\circ)} = s i n ( 2 × 2 0 ∘ ) 4 ( s i n 6 0 ∘ c o s 2 0 ∘ − c o s 6 0 ∘ s i n 2 0 ∘ ) = 4 sin ( 60 ∘ − 20 ∘ ) sin 40 ∘ = \frac{4\sin(60^\circ - 20^\circ)}{\sin 40^\circ} = s i n 4 0 ∘ 4 s i n ( 6 0 ∘ − 2 0 ∘ ) = 4 sin 40 ∘ sin 40 ∘ = 4 = ডানপক্ষ = \frac{4\sin 40^\circ}{\sin 40^\circ} = 4 = \text{ডানপক্ষ} = s i n 4 0 ∘ 4 s i n 4 0 ∘ = 4 = ডানপক্ষ
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, K = cos P + cos R K = \cos P + \cos R K = cos P + cos R P + Q + R = π P + Q + R = \pi P + Q + R = π = K − 1 + cos Q = K - 1 + \cos Q = K − 1 + cos Q = cos P + cos R + cos Q − 1 = \cos P + \cos R + \cos Q - 1 = cos P + cos R + cos Q − 1 = ( cos P + cos R ) − ( 1 − cos Q ) = (\cos P + \cos R) - (1 - \cos Q) = ( cos P + cos R ) − ( 1 − cos Q ) = 2 cos P + R 2 cos P − R 2 − 2 sin 2 Q 2 = 2\cos \frac{P+R}{2} \cos \frac{P-R}{2} - 2\sin^2 \frac{Q}{2} = 2 cos 2 P + R cos 2 P − R − 2 sin 2 2 Q
যেহেতু P + Q + R = π ⟹ P + R 2 = π 2 − Q 2 P + Q + R = \pi \implies \frac{P+R}{2} = \frac{\pi}{2} - \frac{Q}{2} P + Q + R = π ⟹ 2 P + R = 2 π − 2 Q ∴ cos P + R 2 = cos ( π 2 − Q 2 ) = sin Q 2 \therefore \cos \frac{P+R}{2} = \cos\left(\frac{\pi}{2} - \frac{Q}{2}\right) = \sin \frac{Q}{2} ∴ cos 2 P + R = cos ( 2 π − 2 Q ) = sin 2 Q
অতএব,= 2 sin Q 2 cos P − R 2 − 2 sin 2 Q 2 = 2\sin \frac{Q}{2} \cos \frac{P-R}{2} - 2\sin^2 \frac{Q}{2} = 2 sin 2 Q cos 2 P − R − 2 sin 2 2 Q = 2 sin Q 2 [ cos P − R 2 − sin Q 2 ] = 2\sin \frac{Q}{2} \left[\cos \frac{P-R}{2} - \sin \frac{Q}{2}\right] = 2 sin 2 Q [ cos 2 P − R − sin 2 Q ] = 2 sin Q 2 [ cos P − R 2 − cos P + R 2 ] = 2\sin \frac{Q}{2} \left[\cos \frac{P-R}{2} - \cos \frac{P+R}{2}\right] = 2 sin 2 Q [ cos 2 P − R − cos 2 P + R ] = 2 sin Q 2 [ 2 sin P 2 sin R 2 ] = 2\sin \frac{Q}{2} \left[2\sin \frac{P}{2} \sin \frac{R}{2}\right] = 2 sin 2 Q [ 2 sin 2 P sin 2 R ] = 4 sin P 2 sin Q 2 sin R 2 = ডানপক্ষ = 4\sin \frac{P}{2} \sin \frac{Q}{2} \sin \frac{R}{2} = \text{ডানপক্ষ} = 4 sin 2 P sin 2 Q sin 2 R = ডানপক্ষ
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, K = sin Q K = \sin Q K = sin Q cos P + cos R = sin Q \cos P + \cos R = \sin Q cos P + cos R = sin Q 2 cos P + R 2 cos P − R 2 = 2 sin Q 2 cos Q 2 2\cos \frac{P+R}{2} \cos \frac{P-R}{2} = 2\sin \frac{Q}{2} \cos \frac{Q}{2} 2 cos 2 P + R cos 2 P − R = 2 sin 2 Q cos 2 Q
আমরা জানি, P + Q + R = π P + Q + R = \pi P + Q + R = π P Q R PQR P QR ⟹ P + R 2 = π 2 − Q 2 \implies \frac{P+R}{2} = \frac{\pi}{2} - \frac{Q}{2} ⟹ 2 P + R = 2 π − 2 Q ∴ cos P + R 2 = sin Q 2 \therefore \cos \frac{P+R}{2} = \sin \frac{Q}{2} ∴ cos 2 P + R = sin 2 Q
সমীকরণে মান বসিয়ে পাই,2 sin Q 2 cos P − R 2 = 2 sin Q 2 cos Q 2 2\sin \frac{Q}{2} \cos \frac{P-R}{2} = 2\sin \frac{Q}{2} \cos \frac{Q}{2} 2 sin 2 Q cos 2 P − R = 2 sin 2 Q cos 2 Q sin Q 2 ≠ 0 \sin \frac{Q}{2} \neq 0 sin 2 Q = 0 0 < Q < π 0 < Q < \pi 0 < Q < π 2 sin Q 2 2\sin \frac{Q}{2} 2 sin 2 Q cos P − R 2 = cos Q 2 \cos \frac{P-R}{2} = \cos \frac{Q}{2} cos 2 P − R = cos 2 Q ⟹ P − R 2 = Q 2 \implies \frac{P-R}{2} = \frac{Q}{2} ⟹ 2 P − R = 2 Q ⟹ P − R = Q \implies P - R = Q ⟹ P − R = Q ⟹ P = Q + R \implies P = Q + R ⟹ P = Q + R
আবার, P + Q + R = π P + Q + R = \pi P + Q + R = π ⟹ P + P = π \implies P + P = \pi ⟹ P + P = π ⟹ 2 P = π \implies 2P = \pi ⟹ 2 P = π ⟹ P = π 2 = 90 ∘ \implies P = \frac{\pi}{2} = 90^\circ ⟹ P = 2 π = 9 0 ∘
যেহেতু △ P Q R \triangle PQR △ P QR P = 90 ∘ P = 90^\circ P = 9 0 ∘ P Q R PQR P QR