CBSE Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.2
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CBSE Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.2
Exercise 8.2
CBSE Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.2
1.
(i) Evaluate the following
\( \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ} \)
\( \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ} \)
Answer
\( \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ} \)\( =\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \)
\( =\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1 \)
(ii) Evaluate the following
\( 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ} \)
\( 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ} \)
Answer
\( 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ} \)\( =2(1)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2} \)
\( =2+\frac{3}{4}-\frac{3}{4}=2 \)
(iii) Evaluate the following
\( \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}} \)
\( \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}} \)
Answer
\( \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}} \) \(=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}\)
\(=\frac{\sqrt{3}}{\sqrt{2}(2+2 \sqrt{3})}=\frac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}\)
\(=\frac{\sqrt{3}(2 \sqrt{6}-2 \sqrt{2})}{(2 \sqrt{6}+2 \sqrt{2})(2 \sqrt{6}-2 \sqrt{2})}\)
\(=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(2 \sqrt{6})^{2}-(2 \sqrt{2})^{2}}-\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{24-8}=\frac{2 \sqrt{3}(\sqrt{6}-\sqrt{2})}{16}\)
\(=\frac{\sqrt{18}-\sqrt{6}-3 \sqrt{2}-\sqrt{6}}{8}\)
(iv) Evaluate the following
\( \frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}} \)
\( \frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}} \)
Answer
\( \frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}} \)\(=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}\)
\(=\frac{3 \sqrt{3}-4}{\frac{3 \sqrt{3}-4}{2 \sqrt{3}+4}}=\frac{(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)}\)
\(=\frac{(3 \sqrt{3}-4)(3 \sqrt{3}-4)}{(3 \sqrt{3}+4)(3 \sqrt{3}-4)}-\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}\)
\(=\frac{27+16-24 \sqrt{3}}{27-16}-\frac{43-24 \sqrt{3}}{11}\)
(v) Evaluate the following
\( \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} \)
\( \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} \)
Answer
\( \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} \)\( =\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)
\(= \frac{5\left(\frac{1}{4}\right)+\left(\frac{16}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}}\)
\(= \frac{15+64-12}{\frac{4}{4}}=\frac{67}{12}\)
CBSE Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.2
2.
(i) Choose the correct option and justify your choice:
\( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}= \)
(A) \( \sin 60^{\circ} \) (B) \( \cos 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \)
\( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}= \)
(A) \( \sin 60^{\circ} \) (B) \( \cos 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \)
Answer
\( \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \)\( =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\sqrt{3}}{\frac{4}{3}} \)
\( =\frac{6}{4 \sqrt{3}}=\frac{\sqrt{3}}{2} \)
Out of the given alternatives, only \( \sin 60^{\circ}=\frac{\sqrt{3}}{2} \)
Hence, \( (A) \) is correct.
(ii) Choose the correct option and justify your choice:
\( \frac{1-\tan ^{2}}{1+\tan ^{2}} \frac{45^{\circ}}{45^{\circ}}= \)
(A) \( \tan 90^{\circ} \) (B) 1 (C) \( \sin 45^{\circ} \) (D) 0
\( \frac{1-\tan ^{2}}{1+\tan ^{2}} \frac{45^{\circ}}{45^{\circ}}= \)
(A) \( \tan 90^{\circ} \) (B) 1 (C) \( \sin 45^{\circ} \) (D) 0
Answer
\( \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} \)\( =\frac{1-(1)^{2}}{1+(1)^{2}}=\frac{1-1}{1+1}=\frac{0}{2}=0 \)
Hence, (D) is correct.
(iii) Choose the correct option and justify your choice:
\( \sin 2 A=2 \sin A \) is true when \( A= \)
(A) \( 0^{\circ} \) (B) \( 30^{\circ} \) (C) \( 45^{\circ} \) (D) \( 60^{\circ} \)
\( \sin 2 A=2 \sin A \) is true when \( A= \)
(A) \( 0^{\circ} \) (B) \( 30^{\circ} \) (C) \( 45^{\circ} \) (D) \( 60^{\circ} \)
Answer
Out of the given alternatives, only \( A=0^{\circ} \) is correct.As \( \sin 2 \mathrm{~A}=\sin ^{\circ}=0 \)
\( 2 \sin A=2 \sin ^{\circ}=2(0)=0 \)
Hence, \( (A) \) is correct.
(iv) Choose the correct option and justify your choice:
\( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}= \)
(A) \( \cos 60^{\circ} \) (B) \( \sin 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \)
\( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}= \)
(A) \( \cos 60^{\circ} \) (B) \( \sin 60^{\circ} \) (C) \( \tan 60^{\circ} \) (D) \( \sin 30^{\circ} \)
Answer
\( \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} \)\( =\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^{2}}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \)
\(=\sqrt{3}\)
Out of the given alternatives, only \( \tan 60^{\circ}=\sqrt{3} \)
Hence, \( (C) \) is correct.
3. If \( \tan (A+B)=\sqrt{3} \) and \( (A-B)=\frac{1}{\sqrt{3}} \) \( 0^{\circ} < A+B \leq 90^{\circ}, A > B \) find \( A \) and \( B \).
Answer
\(\tan (A+B)=\sqrt{3}\)\(\Rightarrow \tan (A+B)=\tan 60\)
\(\Rightarrow A+B=60 \ldots(1)\)
\(\tan (A-B)=\frac{1}{\sqrt{3}}\)
\(\Rightarrow \tan (A-B)=\frac{1}{\sqrt{3}}\)
\(\Rightarrow A-B=30 \ldots\)(2)
On adding both equations, we obtain
\(2 A=90\)
\(\Rightarrow A=45\)
From equation (1), we obtain
\(45+B=60\)
\(B=15\)
Therefore, \( \angle A=45^{\circ} \) and \( \angle B=15^{\circ} \)
4.
(i) State whether the following are true or false. Justify your answer.
\( \sin (A+B)=\sin A+\sin B \)
\( \sin (A+B)=\sin A+\sin B \)
Answer
\( \sin (A+B)=\sin A+\sin B \)Let \( A=30^{\circ} \) and \( B=60^{\circ} \)
\( \sin (A+B)=\sin \left(30^{\circ}+60^{\circ}\right) \)
\(=\sin 90^{\circ}\)
\(=1\)
\(\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}\)
\(=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\)
Clearly, \( \sin (A+B) \neq \sin A+\sin B \)
Hence, the given statement is false.
(ii) State whether the following are true or false. Justify your answer.
The value of \( \sin \theta \) increases as \( \theta \) increases
The value of \( \sin \theta \) increases as \( \theta \) increases
Answer
The value of \( \sin \theta \) increases as \( \theta \) increases in the interval of \( 0^{\circ} < \theta < 90^{\circ} \) as\(\sin 0^{\circ}=0\)
\(\sin 30^{\circ}=\frac{1}{2}=0.5\)
\(\sin 45^{\circ}=\frac{1}{\sqrt{2}}=0.707\)
\(\sin 60^{\circ}=\frac{\sqrt{3}}{2}=0.866\)
\(\sin 90^{\circ}=\frac{\sqrt{3}}{2}=0.866\)
Hence, the given statement is true.
(iii) State whether the following are true or false. Justify your answer.
The value of \( \cos \theta \) increases as \( \theta \) increases
The value of \( \cos \theta \) increases as \( \theta \) increases
Answer
\( \cos 0^{\circ}=1 \) \(\cos 30^{\circ}=\frac{\sqrt{3}}{2}=0.866\)
\(\cos 45^{\circ}=\frac{1}{\sqrt{2}}=0.707\)
\(\cos 60^{\circ}=\frac{1}{2}=0.5\)
\(\cos 90^{\circ}=0\)
It can be observed that the value of \( \cos \theta \) does not increase in the interval of \( 0^{\circ} < \theta \)
Hence, the given statement is false.
(iv) State whether the following are true or false. Justify your answer.
\( \sin \theta=\cos \theta \) for all values of \( \theta \)
\( \sin \theta=\cos \theta \) for all values of \( \theta \)
Answer
\( \sin \theta=\cos \theta \) for all values of \( \theta \)This is true when \( \theta=45^{\circ} \)
\( \sin 45^{\circ}=\frac{1}{\sqrt{2}} \)
\( \cos 45^{\circ}=\frac{1}{\sqrt{2}} \)
It is not true for all other values of \( \theta \).
Hence, the given statement is false.
(v) State whether the following are true or false. Justify your answer.
\( \cot A \) is not defined for \( A=0^{\circ} \)
\( \cot A \) is not defined for \( A=0^{\circ} \)
Answer
\( \cot A \) is not defined for \( A=0^{\circ} \)\( \cot A=\frac{\cos A}{\sin A} \)
\( \cot 0^{\circ}=\frac{\cos 0^{\circ}}{\sin 0^{\circ}}=\frac{1}{0}= \) undefined
Hence, the given statement is true.
CBSE Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry || CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry solutions Ex 8.2
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