Class 10 Maths Ex 8.2 Assamese Medium Solution introduces students to the deeper concepts of trigonometric identities and their applications. Practicing অনুশীলনী 8.2 Class 10 helps learners strengthen their understanding of trigonometry through step-by-step solutions provided in Assamese. These solutions are designed as per the SEBA Class 10 syllabus to make exam preparation easier and more effective.

Class 10 Maths Ex 8.2 Assamese Medium Solution

Here we provide Class 10 Maths Exercise 8.2 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 8.2 Class 10 Solution

1. তলত দিয়া বিলাকৰ মান উলিওৱা –

(i) \sin 60^\circ cos 30^\circ + \sin 30^\circ cos 60^\circ

(ii) 2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

(iii) \frac{\cos 45^\circ}{\sec 30^\circ + \cosec 30^\circ}

(iv) \frac{\sin 30^\circ + \tan 45^\circ - \cosec 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

(v) \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

সমাধান :

(i) \sin 60^\circ cos 30^\circ + \sin 30^\circ cos 60^\circ

আমি জানো, \sin 60^\circ=\frac{\sqrt{3}}{2} , \cos 30^\circ=\frac{\sqrt{3}}{2} , \sin 30^\circ=\frac{1}{2} , \cos 60^\circ=\frac{1}{2}

এতিয়া,
\sin 60^\circ cos 30^\circ + \sin 30^\circ cos 60^\circ

= (\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}) + (\frac{1}{2}.\frac{1}{2})

= \frac{3}{4} + \frac{1}{4}

= 1

(ii) 2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ

= 2(1)^2 + \left( \frac{\sqrt{3}}{2} \right)^2 - \left( \frac{\sqrt{3}}{2} \right)^2

= 2 \times 1 + \frac{3}{4} - \frac{3}{4}

= 2

(iii) \frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}

= \frac{\tfrac{\sqrt{2}}{2}}{\tfrac{2}{\sqrt{3}} + 2}

= \frac{\tfrac{\sqrt{2}}{2}}{2\left(\tfrac{1}{\sqrt{3}}+1\right)}

= \frac{\sqrt{2}}{2.2(\frac{1+\sqrt{3}}{\sqrt{3}})}

= \frac{\sqrt{2}}{4} \cdot \frac{\sqrt{3}}{1+\sqrt{3}}

= \frac{\sqrt{6}}{4(1+\sqrt{3})}

= \frac{\sqrt{6}}{4(1+\sqrt{3})} \cdot \frac{1-\sqrt{3}}{1-\sqrt{3}}

= \frac{\sqrt{6}(1-\sqrt{3})}{4(1-3)}

= \frac{\sqrt{6}(\sqrt{3}-1)}{8}

= \frac{3\sqrt{2} - \sqrt{6}}{8}

(iv) \frac{\sin 30^\circ + \tan 45^\circ - \cosec 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

আমি জানো, \sin 30^\circ = \tfrac{1}{2} , \tan 45^\circ = 1 , \cosec 60^\circ = \tfrac{2}{\sqrt{3}} , \sec 30^\circ = \tfrac{2}{\sqrt{3}} , \cos 60^\circ = \tfrac{1}{2} আৰু \cot 45^\circ = 1

\frac{\sin 30^\circ + \tan 45^\circ - \cosec 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}

= \frac{\tfrac{1}{2} + 1 - \tfrac{2}{\sqrt{3}}}{\tfrac{2}{\sqrt{3}} + \tfrac{1}{2} + 1}

= \frac{\tfrac{3}{2} - \tfrac{2}{\sqrt{3}}}{\tfrac{3}{2} + \tfrac{2}{\sqrt{3}}}

= \frac{3 - \tfrac{4}{\sqrt{3}}}{3 + \tfrac{4}{\sqrt{3}}}

= \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}

= \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)}

= \frac{9 \cdot 3 + 16 - 24\sqrt{3}}{27 - 16}

= \frac{43 - 24\sqrt{3}}{11}

(v) \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

আমি জানো, \cos 60^\circ = \tfrac{1}{2}, \cos 30^\circ = \tfrac{\sqrt{3}}{2}{}, \sec 30^\circ = \tfrac{2}{\sqrt{3}} , \sec^2 30^\circ = \tfrac{4}{3} , \tan 45^\circ = 1 , \sin 30^\circ = \tfrac{1}{2}

\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}

= \frac{5\left(\tfrac{1}{2}\right)^2 + 4\left(\tfrac{2}{\sqrt{3}}\right)^2 - (1)^2}{\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2}

= \frac{5\cdot\tfrac{1}{4} + 4\cdot\tfrac{4}{3} - 1}{\tfrac{1}{4} + \tfrac{3}{4}}

= \frac{\tfrac{5}{4} + \tfrac{16}{3} - 1}{1}

= \frac{\tfrac{15}{12} + \tfrac{64}{12} - \tfrac{12}{12}}{1}

= \frac{\tfrac{67}{12}}{1}

= \tfrac{67}{12}

2. শুদ্ধ উত্তৰটো বাছি উলিওৱা আৰু তোমাৰ বাছনিৰ যথাৰ্থতা উল্লেখ কৰা।

(i) \dfrac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =

(A) \sin 60^\circ
(B) \cos 60^\circ
(C) \tan 60^\circ
(D) \sin 30^\circ

(ii) \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}=
(A) \tan 90^\circ
(B) 1
(C) \sin 45^\circ
(D) 0

(iii) \sin 2A = 2 \sin A \cos A সত্য়, যেতিয়া A=
(A) 0^\circ
(B) 30^\circ
(C) 45^\circ
(D) 60^\circ

(iv) \frac{2\tan 30^\circ}{1-\tan^2 30^\circ}=
(A) \cos 60^\circ
(B) \sin 60^\circ
(C) \tan 60^\circ
(D) \sin 30^\circ

(i) \frac{2\tan 30^\circ}{1+\tan^2 30^\circ}
উত্তৰঃ (A) \sin 60^\circ

সমাধানঃ

আমি জানো, \tan 30^\circ = \frac{1}{\sqrt{3}}

∴ \frac{2\tan 30^\circ}{1+\tan^2 30^\circ}

= \frac{2\cdot\frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}

= \frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}

= \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}

= \frac{2}{\sqrt{3}}\cdot\frac{3}{4}

= \frac{6}{4\sqrt{3}}

= \frac{3}{2\sqrt{3}}

= \frac{3}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}

= \frac{3\sqrt{3}}{2\cdot 3}

= \frac{\sqrt{3}}{2}

= \frac{\sqrt{3}}{2}=\sin 60^\circ

(ii) \frac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ}=
উত্তৰঃ(D) 0

সমাধানঃ

আমি জানো, \tan 45^\circ = 1

∴ \frac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ}

= \frac{1-1}{1+1}

= \frac{0}{2}

= 0

(iii) \sin 2A = 2 \sin A \cos A সত্য়, যেতিয়া A=
উত্তৰঃ (A)=0^\circ

সমাধানঃ

A ৰ মান বিচাৰিবলৈ বিকল্পবোৰত দিয়া কোণৰ মান এটা এটাকৈ সলনি কৰিলে পোৱা যায়

\sin 2A = 2 \sin A \cos A সত্য়, যেতিয়া A=0 লোৱা হয়।

L.H.S,
\sin 2A
= \sin (2\times 0^\circ)
= \sin 0^\circ
= 0

R.H.S,
2\sin A
= 2\sin 0^\circ
= 2 \times 0
= 0

(iv) \frac{2\tan 30^\circ}{1-\tan^2 30^\circ}=
উত্তৰঃ (C) \tan 60^\circ

সমাধানঃ

আমি জানো, \tan 30^\circ = \frac{1}{\sqrt{3}}

\frac{2 \times\frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^{2}}

= \frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}

= \frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}

= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}

= \frac{2}{\sqrt{3}} \times\frac{3}{2}

= \sqrt {3}

= \tan 60^\circ

3. যদি \tan(A+B) = \sqrt{3} আৰু \tan(A - B) = \frac{1}{\sqrt{3}}, 0^\circ < A+B \leq 90^\circ ; A > B, তেন্তে A আৰু B উলিওৱা।

সমাধান :

\tan(A+B) = \sqrt{3}
⇒ \tan(A+B) = \tan 60^\circ ∵ [ \tan 60^\circ=\sqrt{3} ]
⇒ A+B = 60^\circ \qquad ........(i)

আকৌ, \tan(A-B) = \frac{1}{\sqrt{3}}
⇒ \tan(A-B)= \tan 30^\circ ∵ [ \tan 30^\circ=\frac{1}{\sqrt{3}} ]
⇒ A-B = 30^\circ \qquad ........(ii)

তেনে হলে –
\Rightarrow A+B = 60^\circ
\Rightarrow A-B = 30^\circ

এতিয়া, সমীকৰণ (i) যোগ (ii) —
(A+B) + (A-B) = 60^\circ + 30^\circ
⇒ 2A = 90^\circ
⇒ A = 45^\circ

এতিয়া, (i) নং সমীকৰণত A মান বহুৱাই পাওঁ,
A+B = 60^\circ
⇒ 45^\circ + B = 60^\circ
⇒ B = 60^\circ - 45^\circ
⇒ B= 15^\circ

4. তলত দিয়াবিলাক সত্য নে অসত্য কোৱা। তোমাৰ উত্তৰৰ যুক্তি দাঙি ধৰা –

(i) \sin (A+B) = \sin A + \sin B

(ii) \sin \theta ৰ মান বাঢ়ি যায় যদি \theta ৰ মান বাঢ়ে।

(iii) \cos \theta ৰ মান বাঢ়ি যায় যদি \theta ৰ মান বাঢ়ে।

(iv) \theta ৰ সকলো মানৰ বাবে \sin \theta = \cos \theta

(v) A = 0^\circ ৰ বাবে \cot A সংজ্ঞাবদ্ধ নহয়।

সমাধান :

(i) \sin(A+B) = \sin A + \sin B
→ অসত্য

উত্তৰৰ যুক্তিঃ

ধৰা হল, A = 60^\circ, B = 30^\circ

\sin(A+B) = \sin(60^\circ + 30^\circ) = \sin(90^\circ) = 1

আনহাতে,
\sin A + \sin B = \sin 60^\circ + \sin 30^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3}+1}{2}

∴ L.H.S. \neq R.H.S.

সমাধান :

(ii) \sin \theta ৰ মান বাঢ়ি যায় যদি \theta ৰ মান বাঢ়ে।
→ সত্য

উত্তৰৰ যুক্তিঃ

আমি জানো যে, \sin 0° = 0 ,
\sin 45° = \frac{1}{\sqrt{2}} = 0.7 , (প্ৰায়)
\sin 60° = \frac{\sqrt{3}}{2} = 0.87 , (প্ৰায়)
\sin 90° = 1

ইয়াৰ পৰা বুজা যায় যে \sin \theta ৰ মান বঢ়াৰ লগে লগে \theta ৰ মানো বাঢ়ি গৈ থাকে।

সমাধান :

(iii) \cos \theta ৰ মান বাঢ়ি যায় যদি \theta ৰ মান বাঢ়ে।
→ অসত্য

উত্তৰৰ যুক্তিঃ

আমি জানো যে, \cos 0° = 1
\cos 30° = \frac{\sqrt{3}}{2} = 0.87 , (প্ৰায়)
\cos 45° = \frac{1}{\sqrt{2}} = 0.7 , (প্ৰায়)
\cos 60° = \frac{1}{2} = 0.5
\cos 90° = 0

ইয়াৰ পৰা বুজা যায় যে \cos \theta ৰ মান বঢ়াৰ লগে লগে \theta ৰ মান কমি বাঢ়ি গৈ থাকে।

সমাধান :

(iv) \theta ৰ সকলো মানৰ বাবে \sin \theta = \cos \theta
→ অসত্য

উত্তৰৰ যুক্তিঃ

আমি জানো যে, \sin 30° = \frac{1}{2}, \cos 30° = \frac{\sqrt{3}}{2}
∴ \sin 30° \neq \cos 30°

অৰ্থাৎ, \theta ৰ সকলো মানৰ বাবে \sin \theta \neq \cos \theta

সমাধান :

(v) A = 0^\circ ৰ বাবে \cot A সংজ্ঞাবদ্ধ নহয়।
→ অসত্য

উত্তৰৰ যুক্তিঃ

কাৰণ, \cot 0° = \frac{1}{\tan 0°} = \frac{1}{0} (সংজ্ঞাবদ্ধ নহয়।)

Frequently Asked Questions (FAQ)

Benefits of Class 10 Maths Ex 8.2 Assamese Medium Answer

1. Strengthens Understanding of Trigonometric Identities
By solving Class 10 Maths Ex 8.2 Assamese Medium answers, students can clearly understand how trigonometric identities are formed and applied in real mathematical problems.

2. Improves Accuracy in Solving Problems
The Class 10 Maths Exercise 8.2 Assamese Medium solution provides a step-by-step approach that enhances problem-solving accuracy and logical thinking.

3. Boosts Confidence for SEBA Board Exams
Practicing অনুশীলনী 8.2 Class 10 not only improves speed but also builds confidence for tackling similar questions in the SEBA Class 10 final exams.

4. Easy to Learn in Assamese Medium
Learning Class 10 Maths Ex 8.2 in Assamese ensures students can study in their own language, making trigonometry simpler and more relatable.

Conclusion

The Class 10 Maths Ex 8.2 Assamese Medium solutions provided here will guide students to understand trigonometry concepts clearly and prepare better for exams. By practicing these Class 10 Maths Assamese Medium Exercise 8.2 answers, students can score good marks in SEBA Class 10 Mathematics.