ক) AI-তৈরি নমুনা উত্তর
প্রদত্ত রাশি,tan 42 ∘ tan 78 ∘ cot 6 ∘ cot 66 ∘ \frac{\tan 42^\circ \tan 78^\circ}{\cot 6^\circ \cot 66^\circ} c o t 6 ∘ c o t 6 6 ∘ t a n 4 2 ∘ t a n 7 8 ∘ = tan 6 ∘ tan 66 ∘ tan 42 ∘ tan 78 ∘ [ ∵ 1 cot θ = tan θ ] = \tan 6^\circ \tan 66^\circ \tan 42^\circ \tan 78^\circ \quad \left[\because \frac{1}{\cot \theta} = \tan \theta\right] = tan 6 ∘ tan 6 6 ∘ tan 4 2 ∘ tan 7 8 ∘ [ ∵ c o t θ 1 = tan θ ] = sin 6 ∘ sin 66 ∘ sin 42 ∘ sin 78 ∘ cos 6 ∘ cos 66 ∘ cos 42 ∘ cos 78 ∘ = \frac{\sin 6^\circ \sin 66^\circ \sin 42^\circ \sin 78^\circ}{\cos 6^\circ \cos 66^\circ \cos 42^\circ \cos 78^\circ} = c o s 6 ∘ c o s 6 6 ∘ c o s 4 2 ∘ c o s 7 8 ∘ s i n 6 ∘ s i n 6 6 ∘ s i n 4 2 ∘ s i n 7 8 ∘ = ( 2 sin 66 ∘ sin 6 ∘ ) ( 2 sin 78 ∘ sin 42 ∘ ) ( 2 cos 66 ∘ cos 6 ∘ ) ( 2 cos 78 ∘ cos 42 ∘ ) = \frac{(2 \sin 66^\circ \sin 6^\circ)(2 \sin 78^\circ \sin 42^\circ)}{(2 \cos 66^\circ \cos 6^\circ)(2 \cos 78^\circ \cos 42^\circ)} = ( 2 c o s 6 6 ∘ c o s 6 ∘ ) ( 2 c o s 7 8 ∘ c o s 4 2 ∘ ) ( 2 s i n 6 6 ∘ s i n 6 ∘ ) ( 2 s i n 7 8 ∘ s i n 4 2 ∘ ) = { cos ( 66 ∘ − 6 ∘ ) − cos ( 66 ∘ + 6 ∘ ) } { cos ( 78 ∘ − 42 ∘ ) − cos ( 78 ∘ + 42 ∘ ) } { cos ( 66 ∘ + 6 ∘ ) + cos ( 66 ∘ − 6 ∘ ) } { cos ( 78 ∘ + 42 ∘ ) + cos ( 78 ∘ − 42 ∘ ) } = \frac{\{\cos(66^\circ - 6^\circ) - \cos(66^\circ + 6^\circ)\}\{\cos(78^\circ - 42^\circ) - \cos(78^\circ + 42^\circ)\}}{\{\cos(66^\circ + 6^\circ) + \cos(66^\circ - 6^\circ)\}\{\cos(78^\circ + 42^\circ) + \cos(78^\circ - 42^\circ)\}} = { c o s ( 6 6 ∘ + 6 ∘ ) + c o s ( 6 6 ∘ − 6 ∘ )} { c o s ( 7 8 ∘ + 4 2 ∘ ) + c o s ( 7 8 ∘ − 4 2 ∘ )} { c o s ( 6 6 ∘ − 6 ∘ ) − c o s ( 6 6 ∘ + 6 ∘ )} { c o s ( 7 8 ∘ − 4 2 ∘ ) − c o s ( 7 8 ∘ + 4 2 ∘ )} = ( cos 60 ∘ − cos 72 ∘ ) ( cos 36 ∘ − cos 120 ∘ ) ( cos 72 ∘ + cos 60 ∘ ) ( cos 120 ∘ + cos 36 ∘ ) = \frac{(\cos 60^\circ - \cos 72^\circ)(\cos 36^\circ - \cos 120^\circ)}{(\cos 72^\circ + \cos 60^\circ)(\cos 120^\circ + \cos 36^\circ)} = ( c o s 7 2 ∘ + c o s 6 0 ∘ ) ( c o s 12 0 ∘ + c o s 3 6 ∘ ) ( c o s 6 0 ∘ − c o s 7 2 ∘ ) ( c o s 3 6 ∘ − c o s 12 0 ∘ ) cos 60 ∘ = 1 2 \cos 60^\circ = \frac{1}{2} cos 6 0 ∘ = 2 1 cos 120 ∘ = − 1 2 \cos 120^\circ = -\frac{1}{2} cos 12 0 ∘ = − 2 1 cos 72 ∘ = sin 18 ∘ = 5 − 1 4 \cos 72^\circ = \sin 18^\circ = \frac{\sqrt{5}-1}{4} cos 7 2 ∘ = sin 1 8 ∘ = 4 5 − 1 cos 36 ∘ = 5 + 1 4 \cos 36^\circ = \frac{\sqrt{5}+1}{4} cos 3 6 ∘ = 4 5 + 1 ( cos 60 ∘ − sin 18 ∘ ) ( cos 36 ∘ − cos 120 ∘ ) = ( 1 2 − 5 − 1 4 ) ( 5 + 1 4 + 1 2 ) (\cos 60^\circ - \sin 18^\circ)(\cos 36^\circ - \cos 120^\circ) = \left(\frac{1}{2} - \frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4} + \frac{1}{2}\right) ( cos 6 0 ∘ − sin 1 8 ∘ ) ( cos 3 6 ∘ − cos 12 0 ∘ ) = ( 2 1 − 4 5 − 1 ) ( 4 5 + 1 + 2 1 ) = ( 3 − 5 4 ) ( 3 + 5 4 ) = 9 − 5 16 = 4 16 = 1 4 = \left(\frac{3-\sqrt{5}}{4}\right)\left(\frac{3+\sqrt{5}}{4}\right) = \frac{9-5}{16} = \frac{4}{16} = \frac{1}{4} = ( 4 3 − 5 ) ( 4 3 + 5 ) = 16 9 − 5 = 16 4 = 4 1 ( sin 18 ∘ + cos 60 ∘ ) ( cos 120 ∘ + cos 36 ∘ ) = ( 5 − 1 4 + 1 2 ) ( − 1 2 + 5 + 1 4 ) (\sin 18^\circ + \cos 60^\circ)(\cos 120^\circ + \cos 36^\circ) = \left(\frac{\sqrt{5}-1}{4} + \frac{1}{2}\right)\left(-\frac{1}{2} + \frac{\sqrt{5}+1}{4}\right) ( sin 1 8 ∘ + cos 6 0 ∘ ) ( cos 12 0 ∘ + cos 3 6 ∘ ) = ( 4 5 − 1 + 2 1 ) ( − 2 1 + 4 5 + 1 ) = ( 5 + 1 4 ) ( 5 − 1 4 ) = 5 − 1 16 = 4 16 = 1 4 = \left(\frac{\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}-1}{4}\right) = \frac{5-1}{16} = \frac{4}{16} = \frac{1}{4} = ( 4 5 + 1 ) ( 4 5 − 1 ) = 16 5 − 1 = 16 4 = 4 1 প্রদত্ত রাশির মান = 1 4 1 4 = 1 \text{প্রদত্ত রাশির মান} = \frac{\frac{1}{4}}{\frac{1}{4}} = 1 প্রদত্ত রাশির মান = 4 1 4 1 = 1 1 1 1
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপকের চিত্রে △ S T E \triangle STE △ S T E S S S T T T E E E T E = p TE = p T E = p S E = q SE = q S E = q S T = r ST = r S T = r
ত্রিভুজের সাইন সূত্র অনুসারে আমরা জানি,p sin S = q sin T = r sin E = 2 R \frac{p}{\sin S} = \frac{q}{\sin T} = \frac{r}{\sin E} = 2R s i n S p = s i n T q = s i n E r = 2 R R R R △ S T E \triangle STE △ S T E ∴ p = 2 R sin S \therefore p = 2R \sin S ∴ p = 2 R sin S q = 2 R sin T q = 2R \sin T q = 2 R sin T
এখন, ডানপক্ষ:= p − q p + q cot ( S − T 2 ) = \frac{p - q}{p + q} \cot\left(\frac{S - T}{2}\right) = p + q p − q cot ( 2 S − T ) = 2 R sin S − 2 R sin T 2 R sin S + 2 R sin T cot ( S − T 2 ) = \frac{2R \sin S - 2R \sin T}{2R \sin S + 2R \sin T} \cot\left(\frac{S - T}{2}\right) = 2 R s i n S + 2 R s i n T 2 R s i n S − 2 R s i n T cot ( 2 S − T ) = sin S − sin T sin S + sin T cot ( S − T 2 ) = \frac{\sin S - \sin T}{\sin S + \sin T} \cot\left(\frac{S - T}{2}\right) = s i n S + s i n T s i n S − s i n T cot ( 2 S − T ) = 2 cos ( S + T 2 ) sin ( S − T 2 ) 2 sin ( S + T 2 ) cos ( S − T 2 ) ⋅ cot ( S − T 2 ) = \frac{2 \cos\left(\frac{S + T}{2}\right) \sin\left(\frac{S - T}{2}\right)}{2 \sin\left(\frac{S + T}{2}\right) \cos\left(\frac{S - T}{2}\right)} \cdot \cot\left(\frac{S - T}{2}\right) = 2 s i n ( 2 S + T ) c o s ( 2 S − T ) 2 c o s ( 2 S + T ) s i n ( 2 S − T ) ⋅ cot ( 2 S − T ) = cot ( S + T 2 ) ⋅ tan ( S − T 2 ) ⋅ cot ( S − T 2 ) = \cot\left(\frac{S + T}{2}\right) \cdot \tan\left(\frac{S - T}{2}\right) \cdot \cot\left(\frac{S - T}{2}\right) = cot ( 2 S + T ) ⋅ tan ( 2 S − T ) ⋅ cot ( 2 S − T ) = cot ( S + T 2 ) ⋅ 1 [ ∵ tan θ cot θ = 1 ] = \cot\left(\frac{S + T}{2}\right) \cdot 1 \quad [\because \tan \theta \cot \theta = 1] = cot ( 2 S + T ) ⋅ 1 [ ∵ tan θ cot θ = 1 ] △ S T E \triangle STE △ S T E S + T + E = 180 ∘ S + T + E = 180^\circ S + T + E = 18 0 ∘ S + T = 180 ∘ − E S + T = 180^\circ - E S + T = 18 0 ∘ − E S + T 2 = 90 ∘ − E 2 \frac{S + T}{2} = 90^\circ - \frac{E}{2} 2 S + T = 9 0 ∘ − 2 E
∴ cot ( S + T 2 ) = cot ( 90 ∘ − E 2 ) = tan E 2 \therefore \cot\left(\frac{S + T}{2}\right) = \cot\left(90^\circ - \frac{E}{2}\right) = \tan\frac{E}{2} ∴ cot ( 2 S + T ) = cot ( 9 0 ∘ − 2 E ) = tan 2 E = বামপক্ষ = \text{বামপক্ষ} = বামপক্ষ
অতএব, tan E 2 = p − q p + q cot ( S − T 2 ) \tan\frac{E}{2} = \frac{p - q}{p + q} \cot\left(\frac{S - T}{2}\right) tan 2 E = p + q p − q cot ( 2 S − T )
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে,p 4 + q 4 + r 4 = 2 p 2 ( q 2 + r 2 ) p^4 + q^4 + r^4 = 2p^2(q^2 + r^2) p 4 + q 4 + r 4 = 2 p 2 ( q 2 + r 2 ) p 4 + q 4 + r 4 = 2 p 2 q 2 + 2 p 2 r 2 p^4 + q^4 + r^4 = 2p^2 q^2 + 2p^2 r^2 p 4 + q 4 + r 4 = 2 p 2 q 2 + 2 p 2 r 2 q 4 + r 4 + p 4 − 2 p 2 q 2 − 2 p 2 r 2 = 0 q^4 + r^4 + p^4 - 2p^2 q^2 - 2p^2 r^2 = 0 q 4 + r 4 + p 4 − 2 p 2 q 2 − 2 p 2 r 2 = 0 2 q 2 r 2 2q^2 r^2 2 q 2 r 2 ( q 4 + 2 q 2 r 2 + r 4 ) − 2 p 2 ( q 2 + r 2 ) + p 4 = 2 q 2 r 2 (q^4 + 2q^2 r^2 + r^4) - 2p^2(q^2 + r^2) + p^4 = 2q^2 r^2 ( q 4 + 2 q 2 r 2 + r 4 ) − 2 p 2 ( q 2 + r 2 ) + p 4 = 2 q 2 r 2 ( q 2 + r 2 ) 2 − 2 ( q 2 + r 2 ) p 2 + ( p 2 ) 2 = 2 q 2 r 2 (q^2 + r^2)^2 - 2(q^2 + r^2)p^2 + (p^2)^2 = 2q^2 r^2 ( q 2 + r 2 ) 2 − 2 ( q 2 + r 2 ) p 2 + ( p 2 ) 2 = 2 q 2 r 2 ( q 2 + r 2 − p 2 ) 2 = 2 q 2 r 2 (q^2 + r^2 - p^2)^2 = 2q^2 r^2 ( q 2 + r 2 − p 2 ) 2 = 2 q 2 r 2 q 2 + r 2 − p 2 = ± 2 q r q^2 + r^2 - p^2 = \pm \sqrt{2} qr q 2 + r 2 − p 2 = ± 2 q r 2 q r 2qr 2 q r q 2 + r 2 − p 2 2 q r = ± 2 q r 2 q r \frac{q^2 + r^2 - p^2}{2qr} = \pm \frac{\sqrt{2} qr}{2qr} 2 q r q 2 + r 2 − p 2 = ± 2 q r 2 q r q 2 + r 2 − p 2 2 q r = ± 1 2 \frac{q^2 + r^2 - p^2}{2qr} = \pm \frac{1}{\sqrt{2}} 2 q r q 2 + r 2 − p 2 = ± 2 1
ত্রিভুজের কোসাইন সূত্রানুসারে আমরা জানি, cos S = q 2 + r 2 − p 2 2 q r \cos S = \frac{q^2 + r^2 - p^2}{2qr} cos S = 2 q r q 2 + r 2 − p 2 ∴ cos S = ± 1 2 \therefore \cos S = \pm \frac{1}{\sqrt{2}} ∴ cos S = ± 2 1
(+) চিহ্ন বিবেচনা করে,cos S = 1 2 = cos 45 ∘ \cos S = \frac{1}{\sqrt{2}} = \cos 45^\circ cos S = 2 1 = cos 4 5 ∘ ∴ S = 45 ∘ \therefore S = 45^\circ ∴ S = 4 5 ∘
(-) চিহ্ন বিবেচনা করে,cos S = − 1 2 = − cos 45 ∘ = cos ( 180 ∘ − 45 ∘ ) = cos 135 ∘ \cos S = -\frac{1}{\sqrt{2}} = -\cos 45^\circ = \cos(180^\circ - 45^\circ) = \cos 135^\circ cos S = − 2 1 = − cos 4 5 ∘ = cos ( 18 0 ∘ − 4 5 ∘ ) = cos 13 5 ∘ ∴ S = 135 ∘ \therefore S = 135^\circ ∴ S = 13 5 ∘
অতএব, S = 45 ∘ S = 45^\circ S = 4 5 ∘ 135 ∘ 135^\circ 13 5 ∘
