Mathematics is an important subject for SEBA Class 10 students, and Chapter 8 – Introduction to Trigonometry is one of the most scoring chapters. Here we provide the complete Class 10 Maths Chapter 8 Assamese Medium with step-by-step explanations in simple language. These solutions cover all exercise questions from the SEBA textbook, making it easier for Assamese Medium students to understand trigonometric concepts like sine, cosine, and tangent. With the help of our Class 10 Maths Chapter 8 Assamese Medium solutions, students can strengthen their problem-solving skills, clear doubts, and prepare effectively for the board exams.

Class 10 Maths Chapter 8 Assamese Medium All Exercises

Solving SEBA Class 10 Mathematics textbook exercises helps students build a strong understanding of key concepts, as the questions cover a wide range of problems from all important topics.

Class 10 Maths Ex 8.1 Assamese Medium Solution

Here you can find the complete Class 10 Maths Ex 8.1 Assamese Medium solution with step-by-step answers for every question. These solutions are prepared to help students understand concepts clearly and practice effectively for their SEBA exams.

NCERT Solutions 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 8.1 Class 10 Solution

Class 10 Maths Ex 8.1 Solution

1. \triangle ABC ত্রিভুজৰ B কোণ সমকোণ আৰু AB = 24cm, BC = 7cm হ’লে তলত দিয়াবিলাক উলিওৱাঃ

(i) sinA, cosA
(ii) sinC, cosC

সমাধান :

দিয়া আছে : AB = 24cm, BC = 7cm, \angle B = 90^\circ

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2 = (24)^2 + (7)^2
⇒ AC^2 = 576 + 49 = 625
⇒ AC = \sqrt{625} = 25cm

(i)
\sin A = \frac{BC}{AC} = \frac{7}{25}
\cos A = \frac{AB}{AC} = \frac{24}{25}

(ii)
\sin C = \frac{AB}{AC} = \frac{24}{25}
\cos C = \frac{BC}{AC} = \frac{7}{25}

2. চিত্ৰ 8.13 ৰ পৰা tanP – cotR নিৰ্ণয় কৰা।

সমাধান :

∵ PR = 13cm, PQ = 12cm

PR^2 = PQ^2 + QR^2
⇒ (13)^2 = (12)^2 + (QR)^2
⇒ 169 = 144 + QR^2
⇒ QR^2 = 169 - 144 = 25
⇒ QR = \sqrt{25} = 5cm

\tan P = \frac{QR}{PQ} = \frac{5}{12}
\cot R = \frac{PQ}{QR} = \frac{12}{5}

∴ \tan P - \cot R = \frac{5}{12} - \frac{12}{5} = 0

3. যদি sinA = \frac{3}{4} , তেন্তে cosA আৰু tanA উলিওৱা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ \angle B = 90^\circ

\sin A = \frac{3}{4} = \frac{BC}{AC}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)

∴ BC = 3k, AC = 4k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ (4k)^2 = AB^2 + (3k)^2
⇒ 16k^2 = AB^2 + 9k^2
⇒ AB^2 = 16k^2 - 9k^2 = 7k^2
⇒ AB = \sqrt{7}k

∴ \cos A = \frac{AB}{AC} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4}
∴ \tan A = \frac{BC}{AB} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}}

4. দিয়া আছে যে, 15 cotA = 8, সেই অনুসৰি sinA আৰু secA উলিওৱা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ \angle B = 90^\circ

15 \cot A = 8
⇒ \cot A = \frac{8}{15} = \frac{AB}{BC}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)

AB = 8k, BC = 15k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2 = (8k)^2 + (15k)^2
⇒ AC^2 = 64k^2 + 225k^2
⇒ AC^2 = 289k^2
⇒ AC = \sqrt{289k^2} = 17k

∴ \cos A = \frac{AB}{AC} = \frac{8k}{17k} = \frac{8}{17}
\sec A = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}

5. দিয়া আছে যে, secθ = \frac{13}{12}, আন ত্ৰিকোণমিতিক অনুপাতসমূহৰ গণনা কৰা।

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ, \angle B = 90^\circ, \angle ACB = \theta (ধৰা হল)

⇒ \sec \theta = \frac{13}{12} = \frac{AC}{AB} = k, ধৰা হল (সমানুপাতৰ ভিত্তিত)
AC = 13k, \quad AB = 12k

∵ AC^2 = AB^2 + BC^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ (13k)^2 = (12k)^2 + BC^2
⇒ 169k^2 = 144k^2 + BC^2
⇒ BC^2 = 169k^2 - 144k^2
⇒ BC^2 = 25k^2
⇒ BC = \sqrt{25k^2} = 5k

∴ \sin \theta = \frac{BC}{AC} = \frac{5k}{13k} = \frac{5}{13}
\cos \theta = \frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}
\tan \theta = \frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}
\csc \theta = \frac{AC}{BC} = \frac{13k}{5k} = \frac{13}{5}
\cot \theta = \frac{AB}{BC} = \frac{12k}{5k} = \frac{12}{5}

6. যদি \angle A আৰু \angle B সুক্ষ্মকোণ হয় যাতে \cos A = \cos B , তেতিয়া দেখুৱাব যে \angle A = \angle B

সমাধান :

ধৰা হল, \triangle ABC ত \angle C = 90^\circ

এতিয়া, \cos A = \dfrac{AB}{AC}

আৰু \cos B = \dfrac{BC}{AC}

প্ৰশ্নমতে \cos A = \cos B

⇒ \dfrac{AB}{AC} = \dfrac{BC}{AC}

⇒ AB = AC

∴ ABC এটা সমদ্বিবাহু ত্ৰিভুজ।
∵ আমি জানো যে, ত্ৰিভুজৰ সমান বাহু দুটাৰ বিপৰীত কোণ দুটা সমান।

∴ প্ৰমাণিত হ’ল যে \angle A = \angle B

7. যদি cot\theta=\frac{7}{8} তলৰ মান উলিওৱা –

(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}
(ii) \cot^2\theta

সমাধান :

(i) ধৰা হল, \triangle ABC এটা সমকোণী ত্ৰিভুজ, \angle C = 90^\circ.

\cot \theta = \frac{7}{8}=\frac{BC}{AC}=k ধৰা হল (সমানুপাতৰ ভিত্তিত)
⇒ BC = 7k,\ AC = 8k
⇒ AB^2 = (8k)^2 + (7k)^2
⇒ AB^2= 64k^2 + 49k^2
⇒ AB^2= 113k^2
⇒ AB = \sqrt{113}k

∴ \sin \theta = \frac{AC}{AB} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}
\cos \theta = \frac{BC}{AB} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}

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

= \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2}

= \frac{1-\frac{64}{113}}{1-\frac{49}{113}}

= \frac{\frac{49}{113}}{\frac{64}{113}}

= \frac{49}{64}

(ii) দিয়া আছে, \cot \theta = \frac{7}{8}
⇒ \cot^2\theta = (\frac{7}{8})^2
⇒ \cot^2\theta= \frac{49}{64}

8. যদি 3 \cot A = 4 , তেন্তে \frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A হবনে নহয় পৰীক্ষা কৰা।

সমাধান :
3 \cot A = 4
⇒ \cot A = \frac{4}{3}
⇒ \frac{AB}{BC} = \frac{4}{3}=k , ধৰা হল (সমানুপাতৰ ভিত্তিত)
∴ AB = 4k \quad আৰু \quad BC = 3k

এতিয়া,
AC = \sqrt{(AB)^2 + (BC)^2}
⇒ AC = \sqrt{(4k)^2 + (3k)^2}
⇒ AC = 5k

∵ \sin A = \frac{BC}{AC}
= \frac{3k}{5k}
= \frac{3}{5}
\cos A = \frac{AB}{AC}
= \frac{4k}{5k}
= \frac{4}{5}
\tan A = \frac{BC}{AB}
= \frac{3k}{4k}
= \frac{3}{4}

LHS = \frac{1 - \tan^2 A}{1 + \tan^2 A}
= \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2}
= \frac{1 - \tfrac{9}{16}}{1 + \tfrac{9}{16}}
= \frac{16 - 9}{16 + 9}
= \frac{7}{25}

RHS = \cos^2 A - \sin^2 A
= \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2
= \frac{16}{25} - \frac{9}{25}
= \frac{7}{25}

∵ LHS=RHS
অৰ্থাৎ, \frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A

9. \triangle ABC ত কোনো সমকোণ। যদি, \tan A = \frac{1}{\sqrt{3}} তেন্তে তলৰ মান উলিওৱা –

(i) \sin A \cos C + \cos A \sin C
(ii) \cos A \cos C - \sin A \sin C

সমাধান :

∵ ABC এটা সমকোণী ত্ৰিভুজ, \angle B = 90^\circ, \angle A = \theta.

\tan A = \frac{1}{\sqrt{3}} = \frac{BC}{AB}=k, ধৰা হল (সমানুপাতৰ ভিত্তিত)
∴ BC = k,\ AB = \sqrt{3}k

∵ AC^2 = BC^2 + AB^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ AC^2= k^2 + 3k^2
⇒ AC^2= 4k^2
⇒ AC = 2k

∴ \sin A = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}
\cos A = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}
\sin C = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}
\cos C = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}

(i) \sin A \cos C + \cos A \sin C
= \frac{1}{2}\times\frac{1}{2} + \frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}
= \frac{1}{4} + \frac{3}{4}
= 1

(ii) \cos A \cos C - \sin A \sin C
= \frac{\sqrt{3}}{2}\times\frac{1}{2} - \frac{1}{2}\times\frac{\sqrt{3}}{2}
= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}
= 0

10. \triangle PQR ৰ Q কোণ সমকোণ আৰু PR + QR = 25cm আৰু PQ = 5cm; \cos P আৰু \tan P ৰ মান উলিওৱা।

সমাধান :

দিয়া আছে, PR + QR = 25cm আৰু PQ = 5cm
ধৰা হল, PR = x
\therefore QR = 25-x

PR^2 = QR^2 + PQ^2 (পাইথাগোৰাছ সূত্ৰ)
⇒ x^2 = (25-x)^2 + 25
⇒ x^2 = 625 - 50x + x^2 + 25
⇒ 0 = 650 - 50x
⇒ 50x=650
⇒ x=\frac{650}{50}
⇒ x = 13,
∴ QR = 25-13 = 12

\sin P = \frac{QR}{PR} = \frac{12}{13}
\cos P = \frac{PQ}{PR} = \frac{5}{13}
\tan P = \frac{QR}{PQ} = \frac{12}{5}

Class 10 Maths Ex 8.2 Assamese Medium Solution

Explore the complete Class 10 Maths Ex 8.2 Assamese Medium solutions with accurate step-by-step answers. These Class 10 Maths Chapter 8 Assamese Medium answers are designed to help students master each question and boost their SEBA exam preparation.

অনুশীলনী 8.2 Class 10 Solution

Class 10 Maths Ex 8.2 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।)

Class 10 Maths Ex 8.3 Assamese Medium Solution

Check out the Class 10 Maths Ex 8.3 Assamese Medium solution with clear, easy-to-follow answers. These Class 10 Maths Lesson 8 Assamese Medium solutions guide students to understand each problem and perform better in SEBA exams.

অনুশীলনী 8.3 Class 10 Solution

Class 10 Maths Ex 8.3 Solution

1. মান নিৰ্ণয় কৰা

(i) \frac{\sin 18^\circ}{\cos 72^\circ}

সমাধান :

= \frac{\sin 18^\circ}{\cos (90^\circ - 18^\circ)}

= \frac{\sin 18^\circ}{\sin 18^\circ}

= 1

(ii) \frac{\tan 26^\circ}{\cot 64^\circ}

সমাধান :

= \frac{\tan (90^\circ - 64^\circ)}{\cot 64^\circ}

= \frac{\cot 64^\circ}{\cot 64^\circ}

= 1

(iii) \cos 48^\circ - \sin 42^\circ

সমাধান :

= \cos (90^\circ - 42^\circ) - \sin 42^\circ

= \sin 42^\circ - \sin 42^\circ

= 0

(iv) \csc 31^\circ - \sec 59^\circ

সমাধান :

= \csc (90^\circ - 59^\circ) - \sec 59^\circ

= \sec 59^\circ - \sec 59^\circ

= 0

2. দেখুওৱা যে 

(i) \tan 48^\circ \, \tan 23^\circ \, \tan 42^\circ \, \tan 67^\circ = 1

(ii) \cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0

(i) সমাধান :

LHS = \tan 48^\circ \, \tan 23^\circ \, \tan 42^\circ \, \tan 67^\

= \tan 48^\circ \times \tan 23^\circ \times \tan (90^\circ - 48^\circ) \times \tan (90^\circ - 23^\circ)

= \tan 48^\circ \times \tan 23^\circ \times \cot 48^\circ \times \cot 23^\circ

= \tan 48^\circ \times \tan 23^\circ \times \frac{1}{\tan 48^\circ} \times \frac{1}{\tan 23^\circ}

= 1

∴ RHS = LHS

(ii) সমাধান :

LHS = \cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ

= \cos (90^\circ - 52^\circ) \cos 52^\circ - \sin (90^\circ - 52^\circ) \sin 52^\circ

= \sin 52^\circ \cos 52^\circ - \cos 52^\circ \sin 52^\circ

= 0

∴ RHS = LHS

3. যদি \tan 2A = \cot (A-18^\circ) , য'ত 2A সূক্ষ্মকোণ, তেন্তে A ৰ মান উলিওৱা।

সমাধান :

\tan 2A = \cot (A - 18^\circ)

⇒ \cot (90^\circ - 2A) = \cot (A - 18^\circ)

⇒ 90^\circ - 2A = A - 18^\circ

⇒ 108^\circ = 3A

⇒ A = \frac {108^\circ}{3}

⇒ A = 36^\circ

4. যদি \tan A = \cot B , প্ৰমাণ কৰা যে A + B = 90^\circ .

সমাধান : \tan A = \cot B

⇒ \tan A = \tan (90^\circ - B)

\therefore A = 90^\circ - B

⇒ A + B = 90^\circ

5. যদি \sec 4A = \cosec (A - 20^\circ) , য'ত 4A সূক্ষ্মকোণ, তেন্তে A ৰ মান উলিওৱা।

সমাধান : \sec 4A = \csc (A - 20^\circ)

⇒ \cosec (90^\circ - 4A) = \csc (A - 20^\circ)

⇒ { \cosec (90^\circ - \theta) = \sec \theta }

⇒ 90^\circ - 4A = A - 20^\circ

⇒ 5A = 110^\circ

⇒ A = 22^\circ

6. যদি A, B আৰু C কোণকেইটা ABC ত্ৰিভূজৰ অন্তঃকোণ হয়, তেন্তে দেখুওৱা যে

\sin \left(\frac{B + C}{2}\right) = \cos \left(\frac{A}{2}\right)

সমাধান :

আমি জানো যে, A + B + C = 180^\circ

⇒ B + C = 180^\circ - A

⇒ \frac{B + C}{2} = \frac{180^\circ - A}{2}

⇒ \frac{B + C}{2} = 90^\circ - \frac{A}{2}

⇒ \sin \left(\frac{B + C}{2}\right) = \sin \left(90^\circ - \frac{A}{2}\right)

⇒ \sin \left(\frac{B + C}{2}\right) = \cos \left(\frac{A}{2}\right)

7. \sin 67^\circ + \cos 75^\circ ক 0^\circ আৰু 45^\circ ৰ মাজৰ কোণৰ ত্ৰিকোণমিতিক অনুপাত হিচাপে প্ৰকাশ কৰা।

সমাধান : \sin 67^\circ + \cos 75^\circ

= \sin (90^\circ - 23^\circ) + \cos (90^\circ - 15^\circ)

= \cos 23^\circ + \sin 15^\circ

Class 10 Maths Ex 8.4 Assamese Medium Solution

Class 10 Maths Ex 8.4 Assamese Medium solutions with clear, step-by-step answers. These Class 10 Maths Exercise 8 Assamese Medium solutions help students master all questions and boost their SEBA exam preparation.

অনুশীলনী 8.4 Class 10 Solution

Class 10 Maths Ex 8.4 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

ত্ৰিকোণমিতিৰ পৰিচয় - Class 10 Chapter 8 Assamese Medium

Before learning the basic concepts, it is important to understand the meaning of trigonometry. The word “Trigonometry” comes from the Greek words “tri” (three), “gon” (sides), and “metron” (measure). Trigonometry is an important branch of mathematics that deals with the relationship between the sides and angles of a right-angled triangle. It is widely used in fields like mathematics, physics, astronomy, and architecture.

Right-angled triangles appear in many real-life situations. For example, buildings are constructed at 90-degree angles to ensure stability, and trigonometric principles help calculate heights and distances accurately. In this chapter, students will solve the Class 10 Maths Chapter 8 Assamese Medium exercise questions to build a strong foundation in trigonometry and understand key concepts effectively.

Benefits of Class 10 Maths Chapter 8 Assamese Medium

The Class 10 Maths Chapter 8 Assamese Medium solutions help students build a strong foundation in trigonometry. By studying Class 10 Maths Chapter 8 Assamese Medium answers, students learn basic trigonometric ratios – sine, cosine, and tangent – which are essential for advanced mathematics and real-life applications like engineering, navigation, and architecture. These solutions also improve problem-solving skills, enhance analytical thinking, and prepare students for higher-level trigonometric concepts in future classes.

Summary of Class 10 Maths Chapter 8 Assamese Medium

ত্ৰিকোণমিতিৰ অনুপাত:

1. ABC সমকোণী ত্ৰিভুজত B কোণ সমকোণ হলে—

\sin A = \frac{\text{A কোণৰ বিপৰীত বাহু}}{\text{অতিভুজ}} ,

\cos A = \frac{\text{A কোণৰ সন্নিহিত বাহু}}{\text{অতিভুজ}} ,

\tan A = \frac{\text{A কোণৰ বিপৰীত বাহু}}{\text{A কোণৰ সন্নিহিত বাহু}} ,

2. \cosec A = \frac{1}{\sin A}

3. \sin (90^\circ - A) = \cos A

4. \cos (90^\circ - A) = \sin A

5. \tan (90^\circ - A) = \cot A

6. \cot (90^\circ - A) = \tan A

7. \sec (90^\circ - A) = \cosec A

8. \cosec (90^\circ - A) = \sec A

9. \sin^2 A + \cos^2 A = 1

10. \sec^2 A - \tan^2 A = 1, \quad 0^\circ \leq A < 90^\circ

11. \cosec^2 A = 1 + \cot^2 A, \quad 0^\circ < A \leq 90^\circ

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