ক) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, tan 7 x 2 = 1 3 \tan \frac{7x}{2} = \frac{1}{3} tan 2 7 x = 3 1 sin 2 θ = 2 tan θ 1 + tan 2 θ \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta} sin 2 θ = 1 + t a n 2 θ 2 t a n θ θ = 7 x 2 \theta = \frac{7x}{2} θ = 2 7 x sin 7 x = 2 tan 7 x 2 1 + tan 2 7 x 2 \sin 7x = \frac{2\tan\frac{7x}{2}}{1 + \tan^2\frac{7x}{2}} sin 7 x = 1 + t a n 2 2 7 x 2 t a n 2 7 x = 2 ( 1 3 ) 1 + ( 1 3 ) 2 = 2 3 1 + 1 9 = 2 3 10 9 = 2 3 × 9 10 = 3 5 = \frac{2 \left(\frac{1}{3}\right)}{1 + \left(\frac{1}{3}\right)^2} = \frac{\frac{2}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \frac{2}{3} \times \frac{9}{10} = \frac{3}{5} = 1 + ( 3 1 ) 2 2 ( 3 1 ) = 1 + 9 1 3 2 = 9 10 3 2 = 3 2 × 10 9 = 5 3 sin 7 x = 3 5 \sin 7x = \frac{3}{5} sin 7 x = 5 3
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( x ) = cos x f(x) = \cos x f ( x ) = cos x f ( A ) = cos A f(A) = \cos A f ( A ) = cos A f ( B ) = cos B f(B) = \cos B f ( B ) = cos B f ( π 2 − A ) = cos ( π 2 − A ) = sin A f\left(\frac{\pi}{2} - A\right) = \cos\left(\frac{\pi}{2} - A\right) = \sin A f ( 2 π − A ) = cos ( 2 π − A ) = sin A f ( π 2 − B ) = cos ( π 2 − B ) = sin B f\left(\frac{\pi}{2} - B\right) = \cos\left(\frac{\pi}{2} - B\right) = \sin B f ( 2 π − B ) = cos ( 2 π − B ) = sin B
প্রদত্ত সম্পর্ক:f ( A ) + f ( π 2 − A ) f ( B ) + f ( π 2 − B ) = 1 \frac{f(A) + f\left(\frac{\pi}{2} - A\right)}{f(B) + f\left(\frac{\pi}{2} - B\right)} = 1 f ( B ) + f ( 2 π − B ) f ( A ) + f ( 2 π − A ) = 1 ⟹ cos A + sin A cos B + sin B = 1 \implies \frac{\cos A + \sin A}{\cos B + \sin B} = 1 ⟹ c o s B + s i n B c o s A + s i n A = 1 ⟹ cos A + sin A = cos B + sin B \implies \cos A + \sin A = \cos B + \sin B ⟹ cos A + sin A = cos B + sin B ⟹ sin A − sin B = − ( cos A − cos B ) = cos B − cos A \implies \sin A - \sin B = -(\cos A - \cos B) = \cos B - \cos A ⟹ sin A − sin B = − ( cos A − cos B ) = cos B − cos A
সূত্র প্রয়োগ করে পাই,2 sin ( A − B 2 ) cos ( A + B 2 ) = 2 sin ( A + B 2 ) sin ( A − B 2 ) 2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right) = 2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) 2 sin ( 2 A − B ) cos ( 2 A + B ) = 2 sin ( 2 A + B ) sin ( 2 A − B ) ⟹ 2 sin ( A − B 2 ) cos ( A + B 2 ) − 2 sin ( A + B 2 ) sin ( A − B 2 ) = 0 \implies 2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right) - 2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) = 0 ⟹ 2 sin ( 2 A − B ) cos ( 2 A + B ) − 2 sin ( 2 A + B ) sin ( 2 A − B ) = 0 ⟹ 2 sin ( A − B 2 ) [ cos ( A + B 2 ) − sin ( A + B 2 ) ] = 0 \implies 2\sin\left(\frac{A-B}{2}\right) \left[ \cos\left(\frac{A+B}{2}\right) - \sin\left(\frac{A+B}{2}\right) \right] = 0 ⟹ 2 sin ( 2 A − B ) [ cos ( 2 A + B ) − sin ( 2 A + B ) ] = 0
যেহেতু A ≠ B A \neq B A = B A − B 2 ≠ 0 \frac{A-B}{2} \neq 0 2 A − B = 0 sin ( A − B 2 ) ≠ 0 \sin\left(\frac{A-B}{2}\right) \neq 0 sin ( 2 A − B ) = 0 cos ( A + B 2 ) − sin ( A + B 2 ) = 0 \cos\left(\frac{A+B}{2}\right) - \sin\left(\frac{A+B}{2}\right) = 0 cos ( 2 A + B ) − sin ( 2 A + B ) = 0 ⟹ sin ( A + B 2 ) = cos ( A + B 2 ) \implies \sin\left(\frac{A+B}{2}\right) = \cos\left(\frac{A+B}{2}\right) ⟹ sin ( 2 A + B ) = cos ( 2 A + B ) ⟹ tan ( A + B 2 ) = 1 = tan π 4 \implies \tan\left(\frac{A+B}{2}\right) = 1 = \tan\frac{\pi}{4} ⟹ tan ( 2 A + B ) = 1 = tan 4 π ⟹ A + B 2 = π 4 \implies \frac{A+B}{2} = \frac{\pi}{4} ⟹ 2 A + B = 4 π ⟹ 2 ( A + B ) = π \implies 2(A+B) = \pi ⟹ 2 ( A + B ) = π
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( x ) = cos x f(x) = \cos x f ( x ) = cos x P = 20 ∘ P = 20^\circ P = 2 0 ∘ f ( π 2 − θ ) = cos ( π 2 − θ ) = sin θ f\left(\frac{\pi}{2} - \theta\right) = \cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta f ( 2 π − θ ) = cos ( 2 π − θ ) = sin θ f ( π 2 − θ ) f ( θ ) = sin θ cos θ = tan θ \frac{f\left(\frac{\pi}{2} - \theta\right)}{f(\theta)} = \frac{\sin \theta}{\cos \theta} = \tan \theta f ( θ ) f ( 2 π − θ ) = c o s θ s i n θ = tan θ
বামপক্ষ (LHS):= f ( π 2 − P ) f ( P ) ⋅ f ( π 2 − 2 P ) f ( 2 P ) ⋅ f ( π 2 − 3 P ) f ( 3 P ) ⋅ f ( π 2 − 4 P ) f ( 4 P ) = \frac{f\left(\frac{\pi}{2} - P\right)}{f(P)} \cdot \frac{f\left(\frac{\pi}{2} - 2P\right)}{f(2P)} \cdot \frac{f\left(\frac{\pi}{2} - 3P\right)}{f(3P)} \cdot \frac{f\left(\frac{\pi}{2} - 4P\right)}{f(4P)} = f ( P ) f ( 2 π − P ) ⋅ f ( 2 P ) f ( 2 π − 2 P ) ⋅ f ( 3 P ) f ( 2 π − 3 P ) ⋅ f ( 4 P ) f ( 2 π − 4 P ) = tan P ⋅ tan 2 P ⋅ tan 3 P ⋅ tan 4 P = \tan P \cdot \tan 2P \cdot \tan 3P \cdot \tan 4P = tan P ⋅ tan 2 P ⋅ tan 3 P ⋅ tan 4 P P = 20 ∘ P = 20^\circ P = 2 0 ∘ = tan 20 ∘ ⋅ tan 40 ∘ ⋅ tan 60 ∘ ⋅ tan 80 ∘ = \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 60^\circ \cdot \tan 80^\circ = tan 2 0 ∘ ⋅ tan 4 0 ∘ ⋅ tan 6 0 ∘ ⋅ tan 8 0 ∘ = tan 60 ∘ ⋅ ( tan 20 ∘ tan 40 ∘ tan 80 ∘ ) = \tan 60^\circ \cdot (\tan 20^\circ \tan 40^\circ \tan 80^\circ) = tan 6 0 ∘ ⋅ ( tan 2 0 ∘ tan 4 0 ∘ tan 8 0 ∘ )
আমরা জানি, tan θ tan ( 60 ∘ − θ ) tan ( 60 ∘ + θ ) = tan 3 θ \tan \theta \tan(60^\circ - \theta) \tan(60^\circ + \theta) = \tan 3\theta tan θ tan ( 6 0 ∘ − θ ) tan ( 6 0 ∘ + θ ) = tan 3 θ θ = 20 ∘ \theta = 20^\circ θ = 2 0 ∘ tan 20 ∘ tan ( 60 ∘ − 20 ∘ ) tan ( 60 ∘ + 20 ∘ ) = tan 20 ∘ tan 40 ∘ tan 80 ∘ = tan ( 3 × 20 ∘ ) = tan 60 ∘ \tan 20^\circ \tan(60^\circ - 20^\circ) \tan(60^\circ + 20^\circ) = \tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan(3 \times 20^\circ) = \tan 60^\circ tan 2 0 ∘ tan ( 6 0 ∘ − 2 0 ∘ ) tan ( 6 0 ∘ + 2 0 ∘ ) = tan 2 0 ∘ tan 4 0 ∘ tan 8 0 ∘ = tan ( 3 × 2 0 ∘ ) = tan 6 0 ∘
অতএব,= tan 60 ∘ ⋅ tan 60 ∘ = \tan 60^\circ \cdot \tan 60^\circ = tan 6 0 ∘ ⋅ tan 6 0 ∘ = 3 ⋅ 3 = 3 = \sqrt{3} \cdot \sqrt{3} = 3 = 3 ⋅ 3 = 3 = ডানপক্ষ = \text{ডানপক্ষ} = ডানপক্ষ