NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 || CBSE class 10 maths chapter 8 Ex 8.3
Discover the complete NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (English Medium), with step-by-step explanations for Exercise 8.3. This resource ensures a thorough understanding of Introduction to Trigonometry, helping Class 10 Maths students excel in their exams. If you have any queries regarding the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions Ex 8.3, drop a comment below, and we’ll assist you at the earliest. || NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.3
NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 || CBSE
class 10 maths chapter 8 Ex 8.3
Exercise 8.3
NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 || CBSE class 10 maths chapter 8 Ex 8.3
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(i) Evaluate:
\( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \)
\( \frac{\sin 18^{\circ}}{\cos 72^{\circ}} \)
Answer
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\sin
\left(90^{\circ}-72^{\circ}\right)\)\(=\frac{\cos 72^{\circ}}{\cos 72^{\circ}}=1\)
(ii) Evaluate:
\( \frac{\tan 26^{\circ}}{\cot 64^{\circ}} \)
\( \frac{\tan 26^{\circ}}{\cot 64^{\circ}} \)
Answer
\( \frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan
\left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}} \)\(=\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1\)
(iii) Evaluate:
\( \cos 48^{\circ}-\sin 42^{\circ} \)
\( \cos 48^{\circ}-\sin 42^{\circ} \)
Answer
\( \cos 48^{\circ}-\sin 42^{\circ}=\cos
\left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ} \)\( =\sin 42^{\circ}-\sin 42^{\circ} \)
\( =0 \)
(iv) Evaluate:
\( \operatorname{cosec} 31^{\circ}-\sec 59^{\circ} \)
\( \operatorname{cosec} 31^{\circ}-\sec 59^{\circ} \)
Answer
\(\operatorname{cosec} 31^{\circ}-\sec
59^{\circ}=\operatorname{cosec}\left(90^{\circ}-59^{\circ}\right)-\sec 59^{\circ}\)\(=\sec 59^{\circ}-\sec 59^{\circ}\)
\(=0\)
2.
(i) Show that:
\( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1 \)
\( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1 \)
Answer
\( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan
67^{\circ}\)\(=\tan \left(90^{\circ}-42^{\circ}\right) \tan \left(90^{\circ}-67^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}\)
\(=\cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}\)
\(=\left(\cot 42^{\circ} \tan 42^{\circ}\right)\left(\cot 67^{\circ} \tan 67^{\circ}\right)\)
\(=(1)(1)\)
\(=1\)
NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 || CBSE class 10 maths chapter 8 Ex 8.3
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(ii) Show that:
\( \cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0 \)
\( \cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0 \)
Answer
\( \cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin
52^{\circ}\)\(=\cos \left(90^{\circ}-52^{\circ}\right) \cos \left(90^{\circ}-38^{\circ}\right)-\sin 38^{\circ} \sin 52^{\circ}\)
\(=\sin 52^{\circ} \sin 38^{\circ}-\sin 38^{\circ} \sin 52^{\circ}\)
\(=0\)
3. If \( \tan 2 A=\cot \left(A-18^{\circ}\right) \), where \( 2 A \) is an acute
angle, find the value of \( A \).
Answer
\(\text { Given that, }\)\(\tan 2 A=\cot \left(A-18^{\circ}\right)\)
\(\cot \left(90^{\circ}-2 A\right)=\cot \left(A-18^{\circ}\right)\)
\(90^{\circ}-2 A=A-18^{\circ}\)
\(108^{\circ}=3 A\)
\(A=36^{\circ}\)
4. If \( \tan A=\cot B \), prove that \( A+B=90^{\circ} \)
Answer
Given that,\( \tan A=\cot B \)
\( \tan A=\tan \left(90^{\circ}-B\right) \)
\(A=90^{\circ}-B\)
\(A+B=90^{\circ}\)
5. If \( \sec 4 A=\operatorname{cosec}\left(A-20^{\circ}\right) \), where \( 4 A
\) is
an acute angle, find the value of \( A \).
Answer
\(\text { Given that, }\)\(\sec 4 A=\operatorname{cosec}\left(A-20^{\circ}\right)\)
\(\operatorname{cosec}\left(90^{\circ}-4 A\right)=\operatorname{cosec}\left(A-20^{\circ}\right)\)
\(90^{\circ}-4 A=A-20^{\circ}\)
\(110^{\circ}=5 A\)
\(A=22^{\circ}\)
6. If \( A, B \) Band \( C \) are interior angles of a triangle \( A B C \) then
show
that
\( \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} \)
\( \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} \)
Answer
We know that for a triangle \( A B C \)\(\angle A+\angle B+\angle C=180^{\circ}\)
\(\angle B+\angle C=180^{\circ}-\angle A\)
\(\frac{\angle B+\angle C}{2}=90^{\circ}-\frac{\angle A}{2}\)
\(\sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)\)
\(=\cos \left(\frac{A}{2}\right)\)
7. Express \( \sin 67^{\circ}+\cos 75^{\circ} \) in terms of trigonometric ratios
of
angles between \( 0^{\circ} \) and \( 45^{\circ} \) .
Answer
\( \sin 67^{\circ}+\cos 75^{\circ}\)\( =\sin \left(90^{\circ}-23^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right)\)
\(= \cos 23^{\circ}+\sin 15^{\circ}\)
NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3 || CBSE class 10 maths chapter 8 Ex 8.3
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