The Class 10 Maths Ex 8.4 Assamese Medium solution offers detailed step-by-step explanations to all the questions from অনুশীলনী 8.4. This exercise mainly focuses on solving trigonometric identities and applying them in different types of problems. With these Assamese Medium Class 10 Maths solutions, students can build accuracy and gain confidence in applying formulas correctly.

Class 10 Maths Ex 8.1 Assamese Medium Solution

Here we provide Class 10 Maths Exercise 8.4 Assamese Medium solutions with detailed steps. Each question is solved in Assamese so that SEBA students can learn easily and improve their problem-solving skills.

SEBA Class 10 Maths Chapter 8: Exercises	Total Number of Questions
Class 10 Maths Ex 8.1 Assamese Medium	11
Class 10 Maths Ex 8.2 Assamese Medium	4
Class 10 Maths Ex 8.3 Assamese Medium	7
Class 10 Maths Ex 8.4 Assamese Medium	5

অনুশীলনী 8.4 Class 10 Solution

1. \sin A , \sec A আৰু \tan A এই ত্ৰিকোণমিতিক অনুপাত কেইটাক \cot A ৰ দ্বাৰা প্ৰকাশ কৰা।

সমাধান :

আমি জানো যে, \cosec^2 A - \cot^2 A = 1

⇒ \cosec^2 A = 1 + \cot^2 A

⇒ \csc A)^2 = \cot^2 A + 1

⇒ \left(\frac{1}{\sin A}\right)^2 = \cot^2 A + 1

⇒ (\sin A)^2 = \frac{1}{\cot^2 A + 1}

⇒ \sin A = \pm \frac{1}{\sqrt{\cot^2 A + 1}}

ইয়াৰ পৰা পাওঁ,
\sin A = \frac{1}{\sqrt{\cot^2 A + 1}}

লগতে,
\tan A = \frac{1}{\cot A}

এতিয়া,
\sec^2 A - \tan^2 A = 1

⇒ \sec^2 A = 1 + \tan^2 A

⇒ \sec^2 A = 1 + \frac{1}{\cot^2 A}

⇒ \sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}

⇒ \sec A = \frac{\sqrt{\cot^2 A + 1}}{\cot A}

2. \sec A ৰ সহায়ত ∠A ৰ আন সকলোবিলাক ত্ৰিকোণমিতিক অনুপাত লিখা।

সমাধান :

আমি জানো যে,
\cos^2 A + \sin^2 A = 1

⇒ \sin^2 A = 1 - \cos^2 A

⇒ \sin A = \sqrt{1 - \cos^2 A}

⇒ \sin A = \sqrt{1 - \frac{1}{\sec^2 A}}

⇒ \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}

লগতে, \cos A = \frac{1}{\sec A}

∵ 1 + \tan^2 A = sec^2 A

⇒ \tan^2 A = sec^2 A - 1

⇒ \tan A = \sqrt{\sec^2 A - 1}

∵ \cot A = \frac{1}{\tan A}

⇒ \cot A = \frac{1}{\sqrt{\sec^2 A - 1}}

∵ \cosec A = \frac{1}{\sin A}

⇒ \cosec A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}

3. মান নিৰ্ণয় কৰা 一

(i) \dfrac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ}

(ii) \sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ

(i) সমাধান :
\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ}

= \frac{\sin^2 (90^\circ - 27^\circ) + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 (90^\circ - 17^\circ)}

= \frac{\cos^2 27^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \sin^2 17^\circ}

= \frac{1}{1}

= 1

(ii) সমাধান :
\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ
= \sin (90^\circ - 65^\circ) \cos 65^\circ + \cos (90^\circ - 65^\circ) \sin 65^\circ
= \cos 65^\circ \cos 65^\circ + \sin 65^\circ \sin 65^\circ
= \cos^2 65^\circ + \sin^2 65^\circ
= 1

4. শুদ্ধ উত্তৰটো বাছি উলিওৱা। তোমাৰ বাছনিৰ যথাৰ্থতা সাব্যস্ত কৰা।

(i) 9 \sec^2 A - 9 \tan^2 A =
(A) 1
(B) 9
(C) 8
(D) 0

সমাধান :
9 \sec^2 A - 9 \tan^2 A
= 9(\sec^2 A - \tan^2 A)
= 9 \times 1
= 9
∴ শুদ্ধ বিকল্প: (B)

(ii) (1 + \tan \theta + \sec \theta)(1 + \cot \theta - \cosec \theta) =
(A) 0
(B) 1
(C) 2
(D) -1

সমাধান :
(1 + \tan \theta + \sec \theta)(1 + \cot \theta - \cosec \theta)

= (1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta})(1 + \frac{\cos \theta}{\sin \theta} - \frac {1}{\sin \theta})

= \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)

= \frac{(\cos \theta + \sin \theta)^2 - 1}{\cos \theta \sin \theta}

= \frac{1 + 2\cos \theta \sin \theta - 1}{\cos \theta \sin \theta}

= \frac{2\cos \theta \sin \theta }{\cos \theta \sin \theta}

= 2
∴ শুদ্ধ বিকল্প: (C)

(iii) (\sec A + \tan A)(1 - \sin A) =
(A) \sec A
(B) \sin A
(C) \csc A
(D) \cos A

সমাধান :
(\sec A + \tan A)(1 - \sin A)

= [latex] (\frac{1}{\cos A}+ \frac {\sin A}{\cos A})(1 - \sin A)

= \frac{(1 + \sin A)}{\cos A} (1 - \sin A)

= \frac{1 - \sin^2 A}{\cos A}

= \frac{\cos^2 A}{\cos A}

= \cos A
∴ শুদ্ধ বিকল্প: (D)

(iv) \frac{1 + \tan^2 A}{1 + \cot^2 A} =
(A) \sec^2 A
(B) -1
(C) \cot^2 A
(D) \tan^2 A

সমাধান :
\frac{1 + \tan^2 A}{1 + \cot^2 A}

= \frac{\sec^2 A}{\cosec^2 A}

= \tan^2 A
∴ শুদ্ধ বিকল্প: (D)

5. তলৰ অভেদ কেইটা প্ৰমাণ কৰা যদিহে ইয়াত কোণ বিলাক সূক্ষ্ম কোণ আৰু যাৰ বাবে অভেদ কেইটা সংজ্ঞাবদ্ধ হয় 一

(i) (\cosec \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}

(ii) \dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A

(iii) \dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta

(iv) \dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}

(v) \dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A

(vi) \sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

(vii) \dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta

(viii) (\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A

(ix) (\cosec A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}

(x) \dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A

(i) সমাধান :

LHS = (cosec θ - cot θ)^{2} \

= \left(\frac{1}{\sin θ} - \frac{\cos θ}{\sin θ}\right)^{2}

= \left(\frac{1 - \cos θ}{\sin θ}\right)^{2} \

= \frac{(1 - \cos θ)^{2}}{\sin^{2} θ}

= \frac{(1 - \cos θ)^{2}}{1 - \cos^{2} θ} \

= \frac{(1 - \cos θ)^{2}}{(1 - \cos θ)(1 + \cos θ)}

= \frac{1 - \cos θ}{1 + \cos θ} \
∴ LHS = RHS.

(ii) সমাধান :

LHS = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}

= \frac{\cos^{2} A + (1 + \sin A)^{2}}{\cos A(1 + \sin A)}

= \frac{\cos^{2} A + 1 + \sin^{2} A + 2\sin A}{\cos A(1 + \sin A)}

= \frac{2(1 + \sin A)}{\cos A(1 + \sin A)}

= \frac{2}{\cos A}

= 2 \sec A

= RHS.

(iii) সমাধান :

LHS = \frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ}

= \frac{\frac{\sin θ}{\cos θ}}{1 - \frac{\cos θ}{\sin θ}} + \frac{\frac{\cos θ}{\sin θ}}{1 - \frac{\sin θ}{\cos θ}}

= \frac{\sin^{2} θ}{\cos θ(\sin θ - \cos θ)} - \frac{\cos^{2} θ}{\sin θ(\sin θ - \cos θ)}

= \frac{\sin^{3} θ - \cos^{3} θ}{\cos θ \sin θ (\sin θ - \cos θ)} \

= \frac{(\sin θ - \cos θ)(\sin^{2} θ + \cos^{2} θ + \sin θ \cos θ)}{\cos θ \sin θ (\sin θ - \cos θ)} \

= \frac{1 + \sin θ \cos θ}{\cos θ \sin θ}

= \frac{1}{\cos θ \sin θ} + 1 \

= 1 + \sec θ \cosec θ \
∴ LHS = RHS.

(iv) সমাধান :

LHS = \frac{1 + \sec A}{\sec A}

= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}

= \frac{\cos A + 1}{1}

= 1 + \cos A

RHS = \frac{\sin^{2} A}{1 - \cos A}

= \frac{\sin^{2} A}{1 - \cos A}

= \frac{1 - \cos^{2} A}{(1 - \cos A)}

= \frac{(1 - \cos A)(1 + \cos A)}{(1 - \cos A)}

= 1 + \cos A \
∴ LHS = RHS.

(v) সমাধান :

LHS = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}

= \frac{\frac{\cos A - \sin A +1}{\sin A}}{\frac{\cos A + \sin A - 1}{\sin A}}

= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} \

= \frac{\cot A - 1 + \cosec A}{\cot A + 1 - \cosec A} \

= \frac{(\cosec A + \cot A) - 1}{1 + \cot A - \cosec A} \

= \frac{(\cosec A + \cot A) - (\cosec^{2} A - \cot^{2} A)}{1 + \cot A - \cosec A} \

= \frac{(\cosec A + \cot A)(1 - (\cosec A - \cot A))}{1 + \cot A - \cosec A} \

= \frac{(\cosec A + \cot A)(1 + \cot A - \cosec A)}{1 + \cot A - \cosec A} \

= \cosec A + \cot A

= RHS.

(vi) সমাধান :

LHS = \sqrt{\frac{1 + \sin A}{1 - \sin A}}

= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \

= \sqrt{\frac{(1 + \sin A)^{2}}{1 - \sin^{2} A}}

= \sqrt{\frac{(1 + \sin A)^{2}}{\cos^{2} A}} \

= \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \

= \sec A + \tan A \

= RHS

∴ LHS = RHS.

(vii) সমাধান :

LHS = \frac{\sin θ - 2\sin^{3} θ}{2\cos^{3} θ - \cos θ} \

= \frac{\sin θ(1 - 2\sin^{2} θ)}{\cos θ(2\cos^{2} θ - 1)} \

= \frac{\sin θ(\cos^{2} θ - \sin^{2} θ)}{\cos θ(\cos^{2} θ - \sin^{2} θ)} \

= \tan θ

= RHS.

(viii) সমাধান :

LHS = (\sin A + \cosec A)^{2} + (\cos A + \sec A)^{2} \

= \sin^{2} A + \cosec^{2} A + 2 + \cos^{2} A + \sec^{2} A + 2 \

= 4 + 1 + 1 + \cot^{2} A + 1 + \tan^{2} A \

= 7 + \tan^{2} A + \cot^{2} A

= RHS.

∴ LHS = RHS.

(ix) সমাধান :

LHS = (\cosec A - \sin A)(\sec A - \cos A) \

= \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) \

= \frac{1 - \sin^{2} A}{\sin A} \cdot \frac{1 - \cos^{2} A}{\cos A} \

= \frac{\cos^{2} A}{\sin A} \cdot \frac{\sin^{2} A}{\cos A}

= \sin A \cos A \

RHS = \frac{1}{\tan A + \cot A}

= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} \

= \frac{1}{\frac{\sin^{2} A + \cos^{2} A}{\sin A \cos A}}

= \sin A \cos A \

∴ LHS = RHS.

(x) সমাধান :

LHS = \frac{1 + \tan^{2} A}{1 + \cot^{2} A}

= \frac{\sec^{2} A}{\cosec^{2} A}

= \tan^{2} A

= RHS.

একেদৰে, \left(\frac{1 - \tan A}{1 - \cot A}\right)^{2}

= \left(\frac{\frac{\cos A - \sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^{2} \

= \frac{\sin^{2} A}{\cos^{2} A}

= \tan^{2} A

= RHS

Frequently Asked Questions (FAQ)

Key Features of Class 10 Maths Ex 8.4 Assamese Medium Solution

• Complete Assamese Medium Solutions – Step-by-step answers for every question of অনুশীলনী 8.4.
• Easy to Understand Language – Explained in simple Assamese for better clarity.
• Covers All Important Trigonometric Identities – Helps in understanding the logic behind each step.
• Exam-Oriented Approach – Designed as per the SEBA Class 10 syllabus for scoring high marks.
• Boosts Problem-Solving Skills – Improves speed, accuracy, and confidence in trigonometry.

Conclusion

The Class 10 Maths Ex 8.4 Assamese Medium Solution helps students clearly understand the application of trigonometric identities and formulas. By practicing অনুশীলনী 8.4 step-by-step answers in Assamese, learners can strengthen their conceptual knowledge, improve accuracy, and gain confidence for the SEBA Class 10 final exams.