Question

plz tell fast trigno class 10th
with explanation

Answer 1

hope this will help you

Answer 2

$\boxed{\mathsf{ Solution : }}$

$\mathsf{ Given, \: }$

$\mathtt{ \implies{2 \: \sin \theta \: + \: 3 \: \cos \theta \: = \: 2 \: }} \: \: \: .....( \: 1 \: )$

$\mathtt{ Squaring \: \: both \: \: sides , \: }$

$\mathtt{ \implies( \: 2 \: \sin \theta \: + \: 3 \: \cos \theta \: ) ^{2} \: = \: {2}^{2} } \\ \\ \mathtt{ \implies( \: 2 \: \sin \theta \: )^{2} \: + \: ( \: 3 \: \cos \theta \: )^{2} \: + \: 2 \: \times \: 2 \: \sin \theta \: \times 3 \: \cos \theta \: = \: 4 } \\ \\ \mathtt{ \implies \: 4 \: \sin^{2} \theta \: + \: 9 \: \cos ^{2} \theta \: + \: 12 \: \sin \theta \: \cos \theta \: = \: 4 \: \: .......( \: 2 \: ) }$

$\mathtt{ Let, \: }$

$\mathtt{ \implies{ \: 3 \: \sin \theta \: - \: 2 \: \cos \theta \: = \: x \: }} \: \: \: \: ...... \: ( \: 3 \: )$

$\mathtt{ Squaring \: \: both \: \: sides , \: }$

$\mathtt{ \implies \: ( \: 3 \: \sin \theta \: - \: 2 \: \cos \theta \: ) ^{2} \: = \: {x}^{2} } \\ \\ \mathtt{ \implies \: ( \: 3 \: \sin \theta \: ) ^{2} \: + \: ( \: 2 \: \cos \theta \: )^{2} \: - \: 2 \: \times \: 3 \: \sin \theta \: \times \: 2 \: \cos \theta \: = \: x^{2} } \\ \\ \mathtt{ \implies \: 9 \: \sin^{2} \theta \: + \: 4 \: \cos^{2} \theta \: - \: 12 \: \sin \theta \: \cos \theta \: = \: {x}^{2} \: \: \: \: ......( \: 4 \: )}$

$\mathsf{ Adding \: ( 2 ) \: and \: ( 4 ) \: , \: }$

$\mathtt{ \implies \: 4 \: \sin^{2} \theta \: + \: 9 \: { \sin }^{2} \theta \: + \: 9 \: { \cos }^{2} \theta \: + \: 4 \: { \cos }^{2} \theta \: + \cancel{ \: 12 \: \sin \theta \: cos \: \theta } \: - \cancel{\: 12 \: \sin \theta \: \cos \theta} \: = \: 4 \: + \: {x}^{2} }$

$\mathtt{ \implies \: 13 \: \sin^{2}\theta \: + \: 13 \: \cos ^{2} \theta \: = \: 4 \: + {x}^{2} } \\ \\ \mathtt{ \implies \: 13 \: ( \: \sin^{2} \theta \: + \: \cos^{2} \theta \: ) \: = \: 4 \: + \: {x}^{2} } \\ \\ \mathtt{ \implies \: 13 \: = \: 4 \: + \: {x}^{2} \: \: \quad \quad [ \: ( \sin^{2} \theta \: + \: \cos^{2} \theta \: ) = \: 1 \: ]}$

$\mathtt{ \implies \: 13 \: - \: 4 \: = \: {x}^{2} } \\ \\ \mathtt{ \implies \: 9 \: = \: {x}^{2} } \\ \\ \mathtt{ \implies \: x \: = \: \sqrt{9} } \\ \\ \mathtt{ \implies \: x \: = \: \pm 3 \quad \quad....( \: 5 \: )}$

$\mathsf{ From \: ( 3 ) \: and \: ( 5 ), \: we \: get : \: }$

$\boxed{\mathtt{ \implies( \: 3 \: \sin \theta \: - \: 2 \: \cos \theta \: ) \: = \: \pm3 }}$

$\boxed{\mathsf{ Proved \: \: !! \: }}$