Question

1. If cos
$\alpha + \beta$
= 0, then sin
$\alpha - \beta$
can be reduced
to :​

Answer 1

Given-

$\rm \cos (\alpha + \beta) = 0$

To Evaluate-

$\rm \sin (\alpha - \beta )$

Solution-

$\rm \cos (\alpha + \beta) = 0$

$\rm \implies \cos (\alpha + \beta) = \cos\:90\degree$

$\rm \implies \alpha + \beta = 90\degree$

$\rm \implies \alpha = (90 - \beta)$

Now,

$\implies$$\rm \sin (\alpha - \beta )$

$\implies$$\rm \sin ((90 - \beta) - \beta )$

$\implies$$\rm \sin (90 - 2 \beta )$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\sin(90-\theta) = \cos \: \theta )$

$\implies$$\boxed{ \pink{\cos \:2\beta} }$

Answer 2

Answer:

Solution

verified

Verified by Toppr

Correct option is B)

cos(α+β)=0

⇒cos(α+β)=cos90(∵cos0=0)

⇒α+β=90

⇒α=90−β

sin(α−β)

=sin(90−2β)=cos2β (∵sin(90−θ)=cosθ)

Hence option(B) is the correct answer