CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions 2021
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions 2021
10th Standard CBSE
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Reg.No. :
Maths
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If \(\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } =2,\) find the value of \(\theta\)
(a) -
Prove that \(\sqrt { \sec ^{ 2 }{ \theta } +{ cosec }^{ 2 }\theta } =\tan { \theta } +\cot { \theta } .\)
(a) -
If cos 9 \(\alpha\) = sin \(\alpha\) and value of tan 5 \(\alpha\) is 9 \(\alpha\) < 90°, then the
(a) -
If sin \(\theta\) + 2 cos \(\theta\) = 1. prove that 2 sin \(\theta\) - cos \(\theta\) = 2.
(a) -
If 1 + sin2 \(\theta\) = 3 sin \(\theta\) cos \(\theta\), prove that \(\tan \theta=1 \text { or } \frac{1}{2}\)
(a)
2 Marks
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CBSE 10th Standard Maths Subject Introduction to Trigonometry Ncert Exemplar 2 Marks Questions 2021 Answer Keys
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Given, \(\sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } =2\)
\(\Rightarrow \quad 2\sin ^{ 2 }{ \theta } -(1-\sin ^{ 2 }{ \theta } )=2\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(\Rightarrow \quad 2\sin ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } -1=2\)
\(\Rightarrow \quad 3\sin ^{ 2 }{ \theta } =3\Rightarrow \sin ^{ 2 }{ \theta } =1\Rightarrow \sin { \theta } =1\)
\(\Rightarrow \sin { \theta } =1=\sin { { 90 }^{ 0 } } \quad \left[ \because \sin { { 90 }^{ 0 } } =1 \right] \)
\(\therefore \quad \theta ={ 90 }^{ 0 }\) -
LHS = \(\sqrt { \sec ^{ 2 }{ \theta } +{ cosec }^{ 2 }\theta } =\sqrt { \frac { 1 }{ \cos ^{ 2 }{ \theta } } +\frac { 1 }{ \sin ^{ 2 }{ \theta } } } \)
\(=\sqrt { \frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } } } =\sqrt { \frac { 1 }{ \cos ^{ 2 }{ \theta } \sin ^{ 2 }{ \theta } } } \)
\(=\frac { 1 }{ \sin { \theta } \cos { \theta } } \left[ \therefore \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } \left[ \therefore \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { \sin ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } +\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } \cos { \theta } } =\frac { \sin { \theta } }{ \cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \)
\(=\tan { \theta } +\cot { \theta } \)
= RHS -
cos9 \(\alpha\) = sin \(\alpha\)
\(\Rightarrow \sin \left(90^{\circ}-9 \alpha\right)=\sin \alpha\) \(\left[\because \cos \theta=\sin \left(90^{\circ}-\theta\right)\right]\)
90°- 90 \(\alpha\) = \(\alpha\) \(\left[\because 9 \alpha<90^{\circ} \text { i. e. acute angle }\right]\)
\(\Rightarrow \quad 10 \alpha=90^{\circ} \Rightarrow \alpha=9^{\circ}\)
Hence, find tan 5\(\alpha\)
= 1 -
Given, sin \(\theta\) + 2 cos \(\theta\) = 1
On squaring both sides we get,
\(\sin ^{2} \theta+4 \cos ^{2} \theta+4 \sin \theta \cos \theta=\sin ^{2} \theta+\cos ^{2} \theta\)
\(\Rightarrow \sin ^{2} \theta+4 \cos ^{2} \theta=\sin ^{2} \theta+\cos ^{2} \theta-4 \sin \theta \cos \theta\)
\(\Rightarrow 4 \sin ^{2} \theta+4 \cos ^{2} \theta=4 \sin ^{2} \theta+\cos ^{2} \theta-4 \sin \theta \cos \theta\) [\(\therefore\) adding 3 sin 2 \(\theta\) both sides]
\(\Rightarrow 4\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=(2 \sin \theta-\cos \theta)^{2}\) -
Given, \(1+\sin ^{2} \theta=3 \sin \theta \cos \theta\)
\(\Rightarrow \sec ^{2} \theta+\tan ^{2} \theta=3 \tan \theta\)
\(\Rightarrow 2 \tan ^{2} \theta-3 \tan \theta+1=0\) \(\left[\because \sec ^{2} \theta=1+\tan ^{2} \theta\right]\) [dividing both sides by cos2\(\theta\)]
Let, tan \(\theta\) = x, so above equation becomes 2x2 - 3x + 1. Solve for x = tan \(\theta\) using quadratic formula.
2 Marks
