Question

How to prove this ??

Answer 1

Answer:

LHS = A² + B²

=(SIN ∅ + COS ∅)² + (SIN ∅ - COS ∅)²

=(SIN² ∅ + COS² ∅ + 2SIN∅COS∅) + (SIN²∅ + COS² ∅ -2SIN∅COS∅)

=2(SIN²∅+COS²∅)

SINCE BY PROVEN IDENTITY, SIN²∅ + COS²∅ =1

=2*1

=2

=RHS

Answer 2

Answer:

Given :-

$\sin( \alpha ) + \cos( \alpha ) = a \\ \sin( \alpha ) - \cos( \alpha ) = b$

What to prove =

$a {}^{2} + b {}^{2} = 2$

Solution :-

$\sin ( \alpha ) + \cos( \alpha ) = a \: \\ square \: both \: of \: the \: sides \\ ( \sin( \alpha ) + \cos( \alpha ) ) {}^{2} = a {}^{2} \\ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) + 2 \sin( \alpha ) \times \cos( \alpha ) = a {}^{2} \: \: (equa. \: 1) \\ \sin( \alpha ) - \cos( \alpha ) = b \\ square \: both \: of \: the \: sides \\ we \: get: \\ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) -2 \sin( \alpha ) \cos( \alpha ) = b {}^{2} \: \: (equa. \: 2) \\ now \: add \: equa \: (1) \: \: and \: equa \: (2) \: \\ we \: get :\\ 2 \sin {}^{2} ( \alpha ) + 2 \cos {}^{2} ( \alpha ) = a {}^{2} + b {}^{2} \\ 2( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) ) = a {}^{2} + b {}^{2}$

$∵ \: \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 \\ ∴ \: a {}^{2} + b {}^{2} = 2(1) \\ a {}^{2} + b {}^{2} = 2 \: \: proved$