Question

if acosθ + bsinθ = c , then prove that : ± √a² + b² - c². please answer it if you know ......​

Answer 1

Step-by-step explanation:

so prefer it for the answer

Answer 2

${\huge {\mathfrak \purple{\underline {\underline{\pink{ANSWER }}}}}}$

Step-by-step explanation:

$\rm{we \: have}$

$\bold\sf{(a \cos\theta - b \sin \theta) ^{2}} \\ \bf { + (a \cos\theta + b \sin) \theta ^{2} }$

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

$+ \sf( {a}^{2} \sin ^{2} \theta + b^{2} \cos^{2} \theta + 2a \sin \theta \cos\theta)$

$\sf \implies {a}^{2} ( \cos ^{2} \theta + \sin ^{2} \theta( \sin^{2} \theta + \cos^2 \theta)$

$\sf \implies {a}^{2} + {b}^{2}$

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

${\red{\boxed{\bf\sf[ \because a \cos \theta - b \sin \theta = c ]}}}$

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

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