ক) AI-তৈরি নমুনা উত্তর
দেওয়া আছে,3 tan θ = 1 ⟹ tan θ = 1 3 3\tan\theta = 1 \implies \tan\theta = \frac{1}{3} 3 tan θ = 1 ⟹ tan θ = 3 1
প্রদত্ত রাশি:sin ( π − 4 θ 2 ) = sin ( π 2 − 2 θ ) = cos 2 θ \sin\left(\frac{\pi - 4\theta}{2}\right) = \sin\left(\frac{\pi}{2} - 2\theta\right) = \cos 2\theta sin ( 2 π − 4 θ ) = sin ( 2 π − 2 θ ) = cos 2 θ
আমরা জানি,cos 2 θ = 1 − tan 2 θ 1 + tan 2 θ \cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} cos 2 θ = 1 + t a n 2 θ 1 − t a n 2 θ
tan θ = 1 3 \tan\theta = \frac{1}{3} tan θ = 3 1 cos 2 θ = 1 − ( 1 3 ) 2 1 + ( 1 3 ) 2 = 1 − 1 9 1 + 1 9 = 8 9 10 9 = 8 10 = 4 5 \cos 2\theta = \frac{1 - \left(\frac{1}{3}\right)^2}{1 + \left(\frac{1}{3}\right)^2} = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} = \frac{\frac{8}{9}}{\frac{10}{9}} = \frac{8}{10} = \frac{4}{5} cos 2 θ = 1 + ( 3 1 ) 2 1 − ( 3 1 ) 2 = 1 + 9 1 1 − 9 1 = 9 10 9 8 = 10 8 = 5 4
[বিশেষ দ্রষ্টব্য: যদি মূল প্রশ্ন 3 tan θ = 1 \sqrt{3}\tan\theta = 1 3 tan θ = 1 tan θ = 1 3 ⟹ θ = 30 ∘ \tan\theta = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ tan θ = 3 1 ⟹ θ = 3 0 ∘ cos 2 θ = cos 60 ∘ = 1 2 \cos 2\theta = \cos 60^\circ = \frac{1}{2} cos 2 θ = cos 6 0 ∘ = 2 1
অতএব, নির্ণেয় মান 4 5 \frac{4}{5} 5 4 1 2 \frac{1}{2} 2 1
খ) AI-তৈরি নমুনা উত্তর
উদ্দীপক (i) হতে,T = sec x + tan x T = \sec x + \tan x T = sec x + tan x = 1 cos x + sin x cos x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = c o s x 1 + c o s x s i n x = 1 + sin x cos x = \frac{1 + \sin x}{\cos x} = c o s x 1 + s i n x = cos 2 x 2 + sin 2 x 2 + 2 sin x 2 cos x 2 cos 2 x 2 − sin 2 x 2 = \frac{\cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}} = c o s 2 2 x − s i n 2 2 x c o s 2 2 x + s i n 2 2 x + 2 s i n 2 x c o s 2 x = ( cos x 2 + sin x 2 ) 2 ( cos x 2 − sin x 2 ) ( cos x 2 + sin x 2 ) = \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)} = ( c o s 2 x − s i n 2 x ) ( c o s 2 x + s i n 2 x ) ( c o s 2 x + s i n 2 x ) 2 = cos x 2 + sin x 2 cos x 2 − sin x 2 = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}} = c o s 2 x − s i n 2 x c o s 2 x + s i n 2 x
লব ও হরকে cos x 2 \cos\frac{x}{2} cos 2 x = 1 + tan x 2 1 − tan x 2 = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = 1 − t a n 2 x 1 + t a n 2 x = tan π 4 + tan x 2 1 − tan π 4 tan x 2 = \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} = 1 − t a n 4 π t a n 2 x t a n 4 π + t a n 2 x = tan ( π 4 + x 2 ) = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = tan ( 4 π + 2 x )
অতএব, T = tan ( π 4 + x 2 ) T = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) T = tan ( 4 π + 2 x )
গ) AI-তৈরি নমুনা উত্তর
উদ্দীপক (ii) হতে,M = cos 3 x + cos 3 ( 60 ∘ − x ) + cos 3 ( 60 ∘ + x ) M = \cos^3 x + \cos^3(60^\circ - x) + \cos^3(60^\circ + x) M = cos 3 x + cos 3 ( 6 0 ∘ − x ) + cos 3 ( 6 0 ∘ + x )
উভয়পক্ষকে 4 4 4 4 M = 4 cos 3 x + 4 cos 3 ( 60 ∘ − x ) + 4 cos 3 ( 60 ∘ + x ) 4M = 4\cos^3 x + 4\cos^3(60^\circ - x) + 4\cos^3(60^\circ + x) 4 M = 4 cos 3 x + 4 cos 3 ( 6 0 ∘ − x ) + 4 cos 3 ( 6 0 ∘ + x )
আমরা জানি, 4 cos 3 A = 3 cos A + cos 3 A 4\cos^3 A = 3\cos A + \cos 3A 4 cos 3 A = 3 cos A + cos 3 A ∴ 4 cos 3 ( 60 ∘ − x ) = 3 cos ( 60 ∘ − x ) + cos 3 ( 60 ∘ − x ) = 3 cos ( 60 ∘ − x ) + cos ( 180 ∘ − 3 x ) = 3 cos ( 60 ∘ − x ) − cos 3 x \therefore 4\cos^3(60^\circ - x) = 3\cos(60^\circ - x) + \cos 3(60^\circ - x) = 3\cos(60^\circ - x) + \cos(180^\circ - 3x) = 3\cos(60^\circ - x) - \cos 3x ∴ 4 cos 3 ( 6 0 ∘ − x ) = 3 cos ( 6 0 ∘ − x ) + cos 3 ( 6 0 ∘ − x ) = 3 cos ( 6 0 ∘ − x ) + cos ( 18 0 ∘ − 3 x ) = 3 cos ( 6 0 ∘ − x ) − cos 3 x ∴ 4 cos 3 ( 60 ∘ + x ) = 3 cos ( 60 ∘ + x ) + cos 3 ( 60 ∘ + x ) = 3 cos ( 60 ∘ + x ) + cos ( 180 ∘ + 3 x ) = 3 cos ( 60 ∘ + x ) − cos 3 x \therefore 4\cos^3(60^\circ + x) = 3\cos(60^\circ + x) + \cos 3(60^\circ + x) = 3\cos(60^\circ + x) + \cos(180^\circ + 3x) = 3\cos(60^\circ + x) - \cos 3x ∴ 4 cos 3 ( 6 0 ∘ + x ) = 3 cos ( 6 0 ∘ + x ) + cos 3 ( 6 0 ∘ + x ) = 3 cos ( 6 0 ∘ + x ) + cos ( 18 0 ∘ + 3 x ) = 3 cos ( 6 0 ∘ + x ) − cos 3 x
মানগুলো বসিয়ে পাই,4 M = ( 3 cos x + cos 3 x ) + { 3 cos ( 60 ∘ − x ) − cos 3 x } + { 3 cos ( 60 ∘ + x ) − cos 3 x } 4M = (3\cos x + \cos 3x) + \{3\cos(60^\circ - x) - \cos 3x\} + \{3\cos(60^\circ + x) - \cos 3x\} 4 M = ( 3 cos x + cos 3 x ) + { 3 cos ( 6 0 ∘ − x ) − cos 3 x } + { 3 cos ( 6 0 ∘ + x ) − cos 3 x } = 3 cos x + cos 3 x − 2 cos 3 x + 3 [ cos ( 60 ∘ − x ) + cos ( 60 ∘ + x ) ] = 3\cos x + \cos 3x - 2\cos 3x + 3[\cos(60^\circ - x) + \cos(60^\circ + x)] = 3 cos x + cos 3 x − 2 cos 3 x + 3 [ cos ( 6 0 ∘ − x ) + cos ( 6 0 ∘ + x )] = 3 cos x − cos 3 x + 3 [ 2 cos 60 ∘ cos x ] = 3\cos x - \cos 3x + 3\left[2\cos 60^\circ \cos x\right] = 3 cos x − cos 3 x + 3 [ 2 cos 6 0 ∘ cos x ] = 3 cos x − cos 3 x + 6 ⋅ 1 2 cos x = 3\cos x - \cos 3x + 6 \cdot \frac{1}{2} \cos x = 3 cos x − cos 3 x + 6 ⋅ 2 1 cos x = 3 cos x − cos 3 x + 3 cos x = 3\cos x - \cos 3x + 3\cos x = 3 cos x − cos 3 x + 3 cos x = 6 cos x − cos 3 x = 6\cos x - \cos 3x = 6 cos x − cos 3 x
অতএব, 4 M = 6 cos x − cos 3 x 4M = 6\cos x - \cos 3x 4 M = 6 cos x − cos 3 x