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Home Class 10th Solutions 10th Maths

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

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 8.3. These solutions are based on new NCERT Syllabus.

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

प्रश्न 1.
निम्नलिखित के मान निकालिए :
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
(iii) cos 48° sin 42°
(iv) cosec 31° – sec 590.
हल :
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= 1.
अतः अभीष्ट मान = 1 (एक) है।

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
= 1
अतः अभीष्ट मान 1 (एक) है।

(iii) cos 48° sin 42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42°
= 0
अतः अभीष्ट मान 0 (शून्य) है।

(iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59°
= sec 59° – sec 59°
= 0
अत: अभीष्ट मान 0 (शून्य) है।

प्रश्न 2.
दिखाइए कि:
(i) tan 48°. tan 23°. tan 42° tan 67° = 1
(ii) cos 38°. cos 52° – sin 38°. sin 52° = 0.
हल :
(i) L.H.S. = tan 48°. tan 23°. tan 42° tan 67°
= tan (90° – 42°). tan 23°. tan 42°. tan (90° – 23°)
= cot 42°. tan 23°. tan 42°. cot 23°
= \(\frac{\cos 42^{\circ}}{\sin 42^{\circ}} \cdot \frac{\sin 23^{\circ}}{\cos 23^{\circ}} \cdot \frac{\sin 42^{\circ}}{\cos 42^{\circ}} \cdot \frac{\cos 23^{\circ}}{\sin 23^{\circ}}\)
= 1
= R.H.S.
इति सिद्धम्

(ii) L.H.S. = cos 38°.cos 52° – sin 38°. sin 52°
= cos 38°. cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38°.sin 38° – sin 38°. cos 38°
= cos 38°.sin 38° – cos 38°.sin 38°
= 0
= R.H.S.
इति सिद्धम्

प्रश्न 3.
यदि tan 2A = cot (A – 18°), जहाँ 2A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए।
हल :
चूंकि tan 2A = cot (A – 18°) (दिया हुआ है)
cot (90° – 2A) = cot (A – 18°)
90° – 2A = A – 18°
3A = 90° + 18° = 108°
A = \(\frac { 108 }{ 3 }\)
= 36°
अतः A का अभीष्ट मान = 36° है।

NCERT Solutions

प्रश्न 4.
यदि tan A = cot B, तो सिद्ध कीजिए कि A+ B = 90°.
हल :
चूँकि
⇒ tan A = cot B (दिया है)
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°.
इति सिद्धम्

प्रश्न 5.
यदि sec 4A = cosec (A – 20°) जहाँ 4A एक न्यनूकोण है, तो A का मान ज्ञात कीजिए।
हल :
चूंकि sec 4A = cosec (A – 20°) (दिया है)
⇒ cosec (90° – 4A) = cosec (A – 20°)
⇒ 90° – 4A = A – 20°
⇒ 4A + A = 90° + 20°
⇒ 5A = 110°
⇒ A = \(\frac { 110 }{ 5 }\) = 22°
अतः A का अभीष्ट मान = 22° है।

प्रश्न 6.
यदि A, B और C त्रिभुज ABC के अन्तः कोण हों, तो दिखाइए कि
\(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
हल :
चूंकि
A + B + C = 180° (त्रिभुज के अन्त:कोण हैं)
NCERT Class 10th Maths Solutions Chapter 8 त्रिकोणमिति का परिचय Ex 8.3 1
= cos \(\frac { A }{ 2 }\)
= RHS
इति सिद्धम्

NCERT Solutions

प्रश्न 7.
sin 67° + cos 75° को 0° और 45° के बीच के कोणों के त्रिकोणमितीय अनुपातों के पदों में व्यक्त कीजिए।
हल :
sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
अतः अभीष्ट त्रिकोणमितीय अनुपात के पद cos 23° + sin 15°.

Previous Post

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

Next Post

NCERT Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1

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