Question

If cos (α+β) = 0, then sin(α-β) can be reduced to​

Answer 1

`\text{ cos}( \alpha + \beta ) = 0 \\ \\ \implies \: \alpha + \beta = 90 \degree [\because \text{cos \: 90 \degree} = 0 ] \\ \\ \implies \theta = 90 \degree - \beta \: \: \: \: ......(i)`

`\sin( \alpha - \beta ) \\ = \sin(90 \degree - \beta - \beta ) \: \: \: \: \text{[using \: (i)]} \\ = \sin(90 \degree - 2 \beta ) \\ = \cos(2 \beta ) \: \: \: \: \: \: \: \: \text[ \because \sin(90 \degree - \theta) = \cos( \theta) ]`

Hope This Helps!! :)