Question

If a cos theta +b sin theta = m, a sin theta - b cos theta =n, prove thata2+b2= m2+n2

Answer 1

Given:

$\star \: a \cos \theta \: + b \sin \theta = m - - - -- (1) \\ \\ \star \: a \sin \theta - b \cos \theta = n - - - - - - (2)$

To prove:

$\star \: {a}^{2} + {b}^{2} = {m}^{2} + {n}^{2}$

Solution:

• square both sides and add 1 and 2 : we get ,

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

LHS = RHS

Hence, proved .

Formula :

• (a+b)² = a²+b²+2ab

• (a-b)²= a²+b²-2ab

• sin²@+ cos²@ = 1

Answer 2

Step-by-step explanation:

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