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Home Class 10th Solutions 10th Maths

NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4

by Sudhir
December 28, 2021
in 10th Maths, Class 10th Solutions
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In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4. These solutions are based on new NCERT Syllabus.

NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4

प्रश्न 1.
त्रिकोणमितीय अनुपातों sin A, sec A और tan A को cot A के पदों में व्यक्त कीजिए।
हल:
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 1
अत: cotA के पदों में \(\sin A=\frac{1}{\sqrt{1+\cot ^{2} A}}, \sec A=\frac{\sqrt{1+\cot ^{2} A}}{\cot A}\) एव \(\tan A=\frac{1}{\cot A}\) है।

प्रश्न 2.
∠A के अन्य सभी त्रिकोणमितीय अनुपातों को sec A के पदों में लिखिए।
हल :
मान लीजिए समकोण ∆ABC में ∠B समकोण है और sec A = x = \(\frac { AC }{ AB }\)
यदि कर्ण AC = x तो संलग्न भुजा AB = 1
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 2
अतः sec A के पदों में विभिन्न त्रिकोणमितीय अनुपात हैं :
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 3

प्रश्न 3.
मान निकालिए:
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
(ii) sin 25° cos 65° + cos 25° sin 65°
हल :
(i) \(\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 4
अतः अभीष्ट मान =1

(ii) sin 25°. cos 65° + cos 25°. sin 65°
= sin 25°. cos (90° – 25°) + cos 25°. sin (90° – 25°)
= sin 25°. sin 25° + cos 25°. cos 25°
= sin 25° + cos² 25° = 1 [∴ sin²θ + cos²θ = 1 (सर्वसमिका)]
अतः अभीष्ट मान = 1

NCERT Solutions

प्रश्न 4.
सही विकल्प चुनिए और अपने विकल्प की पुष्टि कीजिए :
(i) 9 sec²A – 9 tan² A बराबर है :
(A) 1
(B) 9
(C) 8
(D) 0.
हल :
सही विकल्प (B) 9 है,
क्योंकि 9 sec² A – 9 tan² A = 9 (sec²A – tan² A)
= 9 x 1
= 9

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) बराबर है :
(A) 0
(B) 1
(C) 2
(D) -1.
हल :
सही विकल्प (C) 2 है, क्योंकि
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 5

(iii) (sec A + tan A) (1 – sin A) बराबर है :
(A) sec A
(B) sin A
(C) coses A
(D) cos A.
हल :
सही विकल्प (D) cos A है,
क्योंकि (sec A + tan A)(1 – sin A)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 6

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) बराबर है :
(A) sec² A
(B) – 1
(C) cot² A
(D) tan² A.
हल :
सही विकल्प (D) tan² A है,
क्योंकि
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 7

प्रश्न 5.
निम्नलिखित सर्वसमिकाएँ सिद्ध कीजिए जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यूनकोण हैं
(i) \((\csc \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}\)
(ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\)
(iii) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \cdot \csc \theta\)
(iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
(v) cosec² A = 1 + cot² A \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A\)
(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A\)
(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta\)
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot² A
(ix) \((\csc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}\)
(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{\cot A}\right)^{2}=\tan ^{2} A\)
हल :
(i) \((\csc \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 8
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 9
LHS = RHS
इति सिद्धम

(ii) \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 10
LHS = RHS
इति सिद्धम

(iii) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \cdot \csc \theta\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 11
LHS = RHS
इति सिद्धम

(iv) \(\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 12
LHS = RHS
इति सिद्धम

(v) cosec² A = 1 + cot² A \(\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 13
LHS = RHS
इति सिद्धम

NCERT Solutions

(vi) \(\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 14
LHS = RHS
इति सिद्धम

(vii) \(\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 15
LHS = RHS
इति सिद्धम

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot² A
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A
= sin² A + cos² A + 2 sin A cosec A + 2 cos A sec A + sec² A + cosec² A
= 1 + 2 + 2 + (1 + tan² A) + (1 + cot² A) [∵ sin² A + cos² A = 1, sin A coses A = 1 एाव cos A sec A = 1 एाव sec² A = 1 + tan² A तथा cosec² A = 1 + cot² A]
= 7 + tan² A + cot² A
= R.H.S.
L.H.S = R.H.S.
इति सिद्धम्

(ix) \((\csc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 16
LHS = RHS
इति सिद्धम

(x) \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{\cot A}\right)^{2}=\tan ^{2} A\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 17
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.4 18
LHS = RHS
इति सिद्धम

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