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Home Class 10th Solutions 10th Maths

NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2

by Sudhir
December 28, 2021
in 10th Maths, Class 10th Solutions
Reading Time: 3 mins read
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NCERT Class 10th Maths Solutions
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In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2. These solutions are based on new NCERT Syllabus.

NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2

प्रश्न 1.
निम्नलिखित के मान निकालिए :
(i) sin 60°. cos 30° + sin 30°. cos 60° (2019)
(ii) 2 tan² 45° + cos² 30° – sin² 60° (2019)
(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\csc 30^{\circ}}\)
(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\csc 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
हल :
(i) sin 60°. cos 30° + sin 30°. cos 60°
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\)
अतः अभीष्ट मान = 1 है।

(ii) 2 tan² 45° + cos² 30° – sin² 60°
\(=2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
\(=2+\frac{3}{4}-\frac{3}{4}\)
अतः अभीष्ट मान = 2 है।

(iii) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\csc 30^{\circ}}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 1
अतः अभीष्ट मान \(\frac{3 \sqrt{2}-\sqrt{6}}{8}\) है।

(iv) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-\csc 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 2
अतः अभीष्ट मान \(\frac{43-24 \sqrt{3}}{11}\) है।

(v) \(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 3
अतः अभीष्ट मान \(\frac { 67 }{ 12 }\) है।

प्रश्न 2.
सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए :
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) = …………
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
हल :
सही विकल्प है (A) sin 60°.
क्योंकि
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 4

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = ……….
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0.
हल :
सही विकल्प (D) 0.
क्योंकि
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 5

(iii) sin 2A = 2 sin A तब सत्य होता है जबकि A बराबर है :
(A) 0°
(B) 30°
(C) 45°
(D) 60°.
हल :
सही विकल्प है (A) 0°.
क्योंकि sin 2A = 2 sin A ⇒ sin 2 x 0° = 2 sin 0
⇒ sin 0° = 2 sin 0° ⇒ 0 = 2 x 0 = 0.

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) बराबर है:
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
हल :
सही विकल्प है (C) tan 60°.
क्योंकि
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.2 6

NCERT Solutions

प्रश्न 3.
यदि tan (A + B) = √3 और tan (A – B) = \(\frac{1}{\sqrt{3}}\), 0° < A + B ≤ 90° ; A > B तो A और B के मान ज्ञात कीजिए।
हल : ∵ tan (A + B) = √3 = tan 60°
⇒ A + B = 60° ….(1)
∵ tan (A – B) = \(\frac{1}{\sqrt{3}}\) = tan 30°
⇒ A – B = 30° …(2)
⇒ 2A = 90°
⇒ A = \(\frac { 90 }{ 2 }\) = 45° [समीकरण (1) + (2) से]
A = 45° का मान समीकरण (1) में रखने पर,
⇒ A + 45° = 60°
⇒ A = 60° – 45° = 15°
अतः A एवं B के अभीष्ट मान हैं क्रमशः 45° एवं 150°.

प्रश्न 4.
बताइए कि निम्नलिखित में कौन-कौन सत्य है या असत्य है? कारण सहित अपने उत्तर की पुष्टि कीजिए:
(i) sin (A + B) = sin A + sin B.
(ii) θ में वृद्धि होने के साथ sin θ के मान में भी वृद्धि होती है।
(iii) θ में वृद्धि होने के साथ cos θ के मान में भी वृद्धि होती है।
(iv) θ के सभी मानों पर sinθ = cosθ.
(v) A = 0° पर cot A परिभाषित नहीं है।
हल :
(i) कथन असत्य है, sin (60° + 30°) = sin 90° = 1
एवं sin 60° + sin 30° = \(\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}\) जो बराबर नहीं है।

(ii) कथन सत्य है, क्योंकि sin 0° = 0, sin 30° = \(\frac { 1 }{ 2 }\), sin 45° = \(\frac{1}{\sqrt{2}}\), sin 60° = \(\frac{\sqrt{3}}{2}\), एवं sin 90° = 1.

(iii) कथन असत्य है, क्योंकि cos 0° = 1, cos 30° = \(\frac{\sqrt{3}}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\), Cos 60° = \(\frac { 1 }{ 2 }\), एवं cos 90° = 0.

(iv) कथन असत्य है, क्योंकि sin 30° = \(\frac { 1 }{ 2 }\) जबकि cos 30° = \(\frac{\sqrt{3}}{2}\).

(v) कथन सत्य है, क्योंकि cot A° = cot 0° = \(\frac{\cos 0^{\circ}}{\sin 0^{\circ}}=\frac{1}{0}\) जो कि अपरिश भाषित है।

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NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.3

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