Question

If a cos-bsin=c then prove that asin+bcos=/a²+b²-c²​

Answer 1

Given :-

$\sf a \cos \theta - b \sin \theta = c$

To Prove :-

$\sf a \sin \theta + b \cos \theta = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$

Solution :-

$\sf a \cos \theta - b \sin \theta = c$

By squaring both sides, we get :-

$: \implies \sf {a}^{2} \cos ^{2} \theta + {b}^{2} { \sin}^{2} \theta - 2ab \cos \theta \sin \theta= {c}^{2}$

$\bullet \bf \: { \cos}^{2} \theta = 1 - { \sin}^{2} \theta$

$\bullet \bf \: { \sin}^{2} = 1 - { \cos}^{2} \theta$

$: \implies \sf {a}^{2} (1 - { \sin}^{2} \theta) + {b}^{2} (1 - { \cos}^{2} \theta) - 2ab \cos \theta \sin \theta = {c}^{2}$

$: \implies \sf {a}^{2} - {a}^{2} \sin ^{2} \theta + {b}^{2} - {b}^{2} \cos ^{2} \theta - 2ab \cos \theta \sin \theta = {c}^{2}$

$: \implies \sf {a}^{2} {sin}^{2} \theta + {b}^{2} { \cos}^{2} \theta + 2ab \cos \theta \sin \theta = {a}^{2} + {b}^{2} - {c}^{2}$

$: \implies \sf (a \sin \theta + b \cos \theta) ^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$: \implies \sf a \sin \theta + b \cos \theta = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$

Hence proved.

Answer 2

Answer:

Given :-

$\sf a \cos \theta - b \sin \theta$

To Prove :-

$\boxed{\sf a \sin \theta + b \cos \theta = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }}$

Solution :-

$\sf a \cos \theta - b \sin \theta$

By squaring both sides, we get :-

$\boxed{\implies \sf {a}^{2} \cos ^{2} \theta + {b}^{2} { \sin}^{2} \theta - 2ab \cos \theta \sin \theta= {c}^{2}:}$

$\bullet \bf \: { \cos}^{2} \theta = 1 - { \sin}^{2} \theta∙$

$\bullet \bf \: { \sin}^{2} = 1 - { \cos}^{2} \theta$

$\boxed { \implies \sf {a}^{2} (1 - { \sin}^{2} \theta) + {b}^{2} (1 - { \cos}^{2} \theta) - 2ab \cos \theta \sin \theta = {c}^{2}:}$

$\boxed{\implies \sf {a}^{2} - {a}^{2} \sin ^{2} \theta + {b}^{2} - {b}^{2} \cos ^{2} \theta - 2ab \cos \theta \sin \theta = {c}^{2}:}$

$\boxed {\implies \sf {a}^{2} {sin}^{2} \theta + {b}^{2} { \cos}^{2} \theta + 2ab \cos \theta \sin \theta = {a}^{2} + {b}^{2} - {c}^{2}:}$

$\boxed{\implies \sf (a \sin \theta + b \cos \theta) ^{2} = {a}^{2} + {b}^{2} - {c}^{2}:}$

$\boxed{\implies \sf a \sin \theta + b \cos \theta = \pm \sqrt{{a}^{2} + {b}^{2} -{c}^{2} }}$

$\pink{Hence \: proved.}$