CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 1 Marks Questions 2021
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 1 Marks Questions 2021
10th Standard CBSE
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Reg.No. :
Maths
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Given that \(\sin { \alpha } =\frac { 1 }{ 2 } \) and \(\cos { \beta } =\frac { 1 }{ 2 } ,\) what is the value of \((\alpha +\beta )?\)
(a)\(sin\ \alpha =\frac { 1 }{ 2 } and\ cos\ \beta =\frac { 1 }{ 2 }\)
\( \\ \Rightarrow d sin\ \alpha =sin\ 30°\left[ \because \ sin\ 30°=\frac { 1 }{ 2 } \right] \ \)
\(\ and\ cos\ \beta \ =\ cos\ 60°\ \left[ \because \ cos\ 60°=\frac { 1 }{ 2 } \right] \)
\(\\ \Rightarrow \ \alpha \ =\ 30°\ and\ \ beta \ = \ 60°\)
\(\\ \therefore \ \alpha \ +\ \beta \ =\ 30°+60°=90°\) -
If \(\cos { 9\alpha } =\sin { \alpha } \) and \(9\alpha <{ 90 }^{ 0 },\) find the value of \(\tan { 5\alpha } .\)
(a)1
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Prove \((\tan { \theta } +2)(2\tan { \theta } +1)=5\tan { \theta } +2\sec ^{ 2 }{ \theta } .\)
(a)LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved. -
Find the value of \(\left[ \frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\sin ^{ 2 }{ { 68 }^{ 0 } } }{ \cos ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\sin { { 27 }^{ 0 } } \cos { { 63 }^{ 0 } } \right] .\)
(a)\(\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\sin ^{ 2 }{ { 68 }^{ 0 } } }{ \cos ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\sin { { 27 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +[\sin { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } }{ [\cos { { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 }] } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \sin { { ({ ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 22 }^{ 0 } } }{ \sin ^{ 2 }{ { 68 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { 1 }{ 1 } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos ^{ 2 }{ { 63 }^{ 0 } } \)
\(=1+1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \) -
If \(\cos { A } +\cos ^{ 2 }{ A } =1,\) find the value of \(\sin ^{ 2 }{ A } +\sin ^{ 4 }{ A } .\)
(a)Given, \(\cos { A } +\cos ^{ 2 }{ A } =1\quad \Rightarrow \cos { A } =1-\cos ^{ 2 }{ A } \)
\(\Rightarrow \quad \cos { A } =\sin ^{ 2 }{ A } \quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(\Rightarrow \quad \cos ^{ 2 }{ A } =\sin ^{ 4 }{ A } \)
On adding \(\sin ^{ 2 }{ A } \) in Eq, (i) both sides, we get
\(\sin ^{ 2 }{ A } +\sin ^{ 4 }{ A } =\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } \)
\(=1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
1 Marks
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 1 Marks Questions 2021 Answer Keys
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\(sin\ \alpha =\frac { 1 }{ 2 } and\ cos\ \beta =\frac { 1 }{ 2 }\)
\( \\ \Rightarrow d sin\ \alpha =sin\ 30°\left[ \because \ sin\ 30°=\frac { 1 }{ 2 } \right] \ \)
\(\ and\ cos\ \beta \ =\ cos\ 60°\ \left[ \because \ cos\ 60°=\frac { 1 }{ 2 } \right] \)
\(\\ \Rightarrow \ \alpha \ =\ 30°\ and\ \ beta \ = \ 60°\)
\(\\ \therefore \ \alpha \ +\ \beta \ =\ 30°+60°=90°\)\(sin\ \alpha =\frac { 1 }{ 2 } and\ cos\ \beta =\frac { 1 }{ 2 }\)
\( \\ \Rightarrow d sin\ \alpha =sin\ 30°\left[ \because \ sin\ 30°=\frac { 1 }{ 2 } \right] \ \)
\(\ and\ cos\ \beta \ =\ cos\ 60°\ \left[ \because \ cos\ 60°=\frac { 1 }{ 2 } \right] \)
\(\\ \Rightarrow \ \alpha \ =\ 30°\ and\ \ beta \ = \ 60°\)
\(\\ \therefore \ \alpha \ +\ \beta \ =\ 30°+60°=90°\) -
1
1
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LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved. -
\(\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\sin ^{ 2 }{ { 68 }^{ 0 } } }{ \cos ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\sin { { 27 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +[\sin { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } }{ [\cos { { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 }] } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \sin { { ({ ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 22 }^{ 0 } } }{ \sin ^{ 2 }{ { 68 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { 1 }{ 1 } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos ^{ 2 }{ { 63 }^{ 0 } } \)
\(=1+1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)\(\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\sin ^{ 2 }{ { 68 }^{ 0 } } }{ \cos ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\sin { { 27 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +[\sin { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } }{ [\cos { { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 }] } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \sin { { ({ ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 22 }^{ 0 } } }{ \sin ^{ 2 }{ { 68 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { 1 }{ 1 } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos ^{ 2 }{ { 63 }^{ 0 } } \)
\(=1+1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \) -
Given, \(\cos { A } +\cos ^{ 2 }{ A } =1\quad \Rightarrow \cos { A } =1-\cos ^{ 2 }{ A } \)
\(\Rightarrow \quad \cos { A } =\sin ^{ 2 }{ A } \quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(\Rightarrow \quad \cos ^{ 2 }{ A } =\sin ^{ 4 }{ A } \)
On adding \(\sin ^{ 2 }{ A } \) in Eq, (i) both sides, we get
\(\sin ^{ 2 }{ A } +\sin ^{ 4 }{ A } =\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } \)
\(=1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)Given, \(\cos { A } +\cos ^{ 2 }{ A } =1\quad \Rightarrow \cos { A } =1-\cos ^{ 2 }{ A } \)
\(\Rightarrow \quad \cos { A } =\sin ^{ 2 }{ A } \quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(\Rightarrow \quad \cos ^{ 2 }{ A } =\sin ^{ 4 }{ A } \)
On adding \(\sin ^{ 2 }{ A } \) in Eq, (i) both sides, we get
\(\sin ^{ 2 }{ A } +\sin ^{ 4 }{ A } =\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } \)
\(=1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
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