ক) AI-তৈরি নমুনা উত্তর
বামপক্ষ = cos 75 ∘ = \cos 75^\circ = cos 7 5 ∘ = cos ( 45 ∘ + 30 ∘ ) = \cos (45^\circ + 30^\circ) = cos ( 4 5 ∘ + 3 0 ∘ ) = cos 45 ∘ cos 30 ∘ − sin 45 ∘ sin 30 ∘ = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ = cos 4 5 ∘ cos 3 0 ∘ − sin 4 5 ∘ sin 3 0 ∘ = 1 2 ⋅ 3 2 − 1 2 ⋅ 1 2 = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = 2 1 ⋅ 2 3 − 2 1 ⋅ 2 1 = 3 − 1 2 2 = \frac{\sqrt{3}-1}{2\sqrt{2}} = 2 2 3 − 1 2 \sqrt{2} 2 = ( 3 − 1 ) 2 2 2 ⋅ 2 = 6 − 2 4 = = \frac{(\sqrt{3}-1)\sqrt{2}}{2\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{6}-\sqrt{2}}{4} = = 2 2 ⋅ 2 ( 3 − 1 ) 2 = 4 6 − 2 =
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, sin θ = 3 5 \sin\theta = \frac{3}{5} sin θ = 5 3 π 2 < θ < π \frac{\pi}{2} < \theta < \pi 2 π < θ < π θ \theta θ sin θ \sin\theta sin θ csc θ \csc\theta csc θ cos θ , tan θ , cot θ \cos\theta, \tan\theta, \cot\theta cos θ , tan θ , cot θ
cos θ = − 1 − sin 2 θ = − 1 − ( 3 5 ) 2 = − 1 − 9 25 = − 16 25 = − 4 5 \cos\theta = -\sqrt{1 - \sin^2\theta} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5} cos θ = − 1 − sin 2 θ = − 1 − ( 5 3 ) 2 = − 1 − 25 9 = − 25 16 = − 5 4 tan θ = sin θ cos θ = 3 5 − 4 5 = − 3 4 \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} tan θ = c o s θ s i n θ = − 5 4 5 3 = − 4 3 cot θ = 1 tan θ = − 4 3 \cot\theta = \frac{1}{\tan\theta} = -\frac{4}{3} cot θ = t a n θ 1 = − 3 4 csc θ = 1 sin θ = 5 3 \csc\theta = \frac{1}{\sin\theta} = \frac{5}{3} csc θ = s i n θ 1 = 3 5
আমরা জানি, cos ( − θ ) = cos θ \cos(-\theta) = \cos\theta cos ( − θ ) = cos θ csc ( − θ ) = − csc θ \csc(-\theta) = -\csc\theta csc ( − θ ) = − csc θ
প্রদত্ত রাশি = cot θ + cos ( − θ ) csc ( − θ ) + tan θ = \frac{\cot\theta + \cos(-\theta)}{\csc(-\theta) + \tan\theta} = c s c ( − θ ) + t a n θ c o t θ + c o s ( − θ ) = cot θ + cos θ − csc θ + tan θ = \frac{\cot\theta + \cos\theta}{-\csc\theta + \tan\theta} = − c s c θ + t a n θ c o t θ + c o s θ = − 4 3 + ( − 4 5 ) − 5 3 + ( − 3 4 ) = \frac{-\frac{4}{3} + \left(-\frac{4}{5}\right)}{-\frac{5}{3} + \left(-\frac{3}{4}\right)} = − 3 5 + ( − 4 3 ) − 3 4 + ( − 5 4 ) = − 4 3 − 4 5 − 5 3 − 3 4 = \frac{-\frac{4}{3} - \frac{4}{5}}{-\frac{5}{3} - \frac{3}{4}} = − 3 5 − 4 3 − 3 4 − 5 4 = − 20 − 12 15 − 20 − 9 12 = \frac{\frac{-20 - 12}{15}}{\frac{-20 - 9}{12}} = 12 − 20 − 9 15 − 20 − 12 = − 32 15 − 29 12 = \frac{-\frac{32}{15}}{-\frac{29}{12}} = − 12 29 − 15 32 = 32 15 × 12 29 = \frac{32}{15} \times \frac{12}{29} = 15 32 × 29 12 = 32 × 4 5 × 29 = 128 145 = \frac{32 \times 4}{5 \times 29} = \frac{128}{145} = 5 × 29 32 × 4 = 145 128
নির্ণেয় মান: 128 145 \frac{128}{145} 145 128
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, A = π 12 = 180 ∘ 12 = 15 ∘ A = \frac{\pi}{12} = \frac{180^\circ}{12} = 15^\circ A = 12 π = 12 18 0 ∘ = 1 5 ∘
বামপক্ষ = tan A tan 3 A tan 5 A tan 7 A tan 11 A = \tan A \tan 3A \tan 5A \tan 7A \tan 11A = tan A tan 3 A tan 5 A tan 7 A tan 11 A tan 3 A = tan ( 3 × 15 ∘ ) = tan 45 ∘ = 1 \tan 3A = \tan (3 \times 15^\circ) = \tan 45^\circ = 1 tan 3 A = tan ( 3 × 1 5 ∘ ) = tan 4 5 ∘ = 1 tan 5 A = tan ( 5 × 15 ∘ ) = tan 75 ∘ \tan 5A = \tan (5 \times 15^\circ) = \tan 75^\circ tan 5 A = tan ( 5 × 1 5 ∘ ) = tan 7 5 ∘ tan 7 A = tan ( 7 × 15 ∘ ) = tan 105 ∘ = tan ( 180 ∘ − 75 ∘ ) = − tan 75 ∘ = − tan 5 A \tan 7A = \tan (7 \times 15^\circ) = \tan 105^\circ = \tan (180^\circ - 75^\circ) = -\tan 75^\circ = -\tan 5A tan 7 A = tan ( 7 × 1 5 ∘ ) = tan 10 5 ∘ = tan ( 18 0 ∘ − 7 5 ∘ ) = − tan 7 5 ∘ = − tan 5 A tan 11 A = tan ( 11 × 15 ∘ ) = tan 165 ∘ = tan ( 180 ∘ − 15 ∘ ) = − tan 15 ∘ = − tan A \tan 11A = \tan (11 \times 15^\circ) = \tan 165^\circ = \tan (180^\circ - 15^\circ) = -\tan 15^\circ = -\tan A tan 11 A = tan ( 11 × 1 5 ∘ ) = tan 16 5 ∘ = tan ( 18 0 ∘ − 1 5 ∘ ) = − tan 1 5 ∘ = − tan A
মানগুলো বসিয়ে পাই,= tan A ⋅ 1 ⋅ tan 5 A ⋅ ( − tan 5 A ) ⋅ ( − tan A ) = \tan A \cdot 1 \cdot \tan 5A \cdot (-\tan 5A) \cdot (-\tan A) = tan A ⋅ 1 ⋅ tan 5 A ⋅ ( − tan 5 A ) ⋅ ( − tan A ) = tan 2 A ⋅ tan 2 5 A = \tan^2 A \cdot \tan^2 5A = tan 2 A ⋅ tan 2 5 A = ( tan 15 ∘ ⋅ tan 75 ∘ ) 2 = (\tan 15^\circ \cdot \tan 75^\circ)^2 = ( tan 1 5 ∘ ⋅ tan 7 5 ∘ ) 2 = { tan 15 ∘ ⋅ tan ( 90 ∘ − 15 ∘ ) } 2 = \{\tan 15^\circ \cdot \tan (90^\circ - 15^\circ)\}^2 = { tan 1 5 ∘ ⋅ tan ( 9 0 ∘ − 1 5 ∘ ) } 2 = ( tan 15 ∘ ⋅ cot 15 ∘ ) 2 = (\tan 15^\circ \cdot \cot 15^\circ)^2 = ( tan 1 5 ∘ ⋅ cot 1 5 ∘ ) 2 = ( 1 ) 2 = (1)^2 = ( 1 ) 2 = 1 = 1 = 1 = = =