Question

if, sinA + cosecA = 2

So prove that,
sin(A-60°) = 1/2​

Answer 1

Answer:

this is your solution for the following information

Answer 2

$\huge\mathfrak\red{Answer}$

$\sin \alpha + \csc \alpha = 2 \\ \\ \sin \alpha + \frac{1}{ \sin \alpha } = 2 \\ \\ \frac{ {sin}^{2} \alpha + 1 }{ \sin \alpha } = 2 \\ \\ { \sin}^{2} \alpha + 1 = 2 \sin \alpha \\ \\ {sin}^{2} \alpha + 1 - 2 \sin \alpha = 0 \\ \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ so \: \\ \\ ( {sin}^{2} \alpha + 1 - 2 \sin \alpha ) = {( \sin \alpha - 1)}^{2} \\ \\ {( \sin \alpha - 1)}^{2} = 0 \\ \\ \sin \alpha - 1 = 0 \\ \\ sin \alpha = 1 \\ \\ \alpha = 90 \\ \\ \sin( \alpha - 60) = \frac{1}{2} \\ \\ \sin(90 - 60) = \frac{1}{2} \\ \\ \sin(30) = \frac{1}{2} \\ \\ \frac{1}{2} = \frac{1}{2}$