Question

Please solve this .............​

Answer 1

Answer:

D

Step-by-step explanation:

Note that I will be using a, b, and y instead of their Greek counterparts.

First, we know that sin(c+d) = sin(c)cos(d) + cos(d)sin(a) and sin(c-d) = sin(c)cos(d) - cos(d)sin(a). We can apply these here to get

sin(b+y) = sin(b)cos(y) + cos(b)sin(y)

sin(b-y) = sin(b)cos(y) - cos(b)sin(y)

sin(a+y) = sin(a)cos(y) + cos(a)sin(y)

sin(a-y) = sin(a)cos(y) - cos(a)sin(y)

Plugging these into our equation, we get

(sin(b)cos(y) + cos(b)sin(y) - (sin(b)cos(y) - cos(b)sin(y)))

/

(sin(a)cos(y) + cos(a)sin(y)-(sin(a)cos(y) - cos(a)sin(y)))

=

2cos(b)sin(y) / 2cos(a)sin(y)

= cos(b)sin(y)/cos(a)sin(y)

= cos(b)/cos(a)

Next, we can see that a+b = π, so we can subtract b from both sides to get a = π -b. Plugging that in for a in our equation, we get

cos(b) / cos(π-b)

After that, we know that cos(c-d) = cos(c)cos(d) - sin(c)sin(d). Plugging that in here, we get

cos(π-b) = cos(π)cos(b) - sin(π)sin(b)

= -cos(b) + 0

= -cos(b)

Plugging that back into our equation, we get

cos(b) / -cos(b) = -1