sinB + cosB cotB = cscB
cscB - sinB
= 1/sinB - sinB
= (1 - sinB*sinB)/sinB
= (1 - sin^2B)/sinB
= cos^2B/sinB
= cosB*(cosB/sinB)
= cosB(cotB)
To prove the statement sin(B) + cos(B) cot(B) = csc(B), we can start by manipulating each side of the equation separately using trigonometric identities.
First, let's look at the left side of the equation:
sin(B) + cos(B) cot(B)
We can rewrite cot(B) as cos(B) / sin(B) using the identity cot(B) = cos(B) / sin(B):
sin(B) + cos(B) * (cos(B) / sin(B))
Now, let's simplify the expression further by multiplying cos(B) with the terms inside the parentheses:
sin(B) + (cos²(B) / sin(B))
Next, we can simplify cos²(B) as 1 - sin²(B) using the Pythagorean identity cos²(B) = 1 - sin²(B):
sin(B) + ((1 - sin²(B)) / sin(B))
Now, let's further simplify the expression by dividing each term by sin(B):
(sin(B) * sin(B) / sin(B)) + (1 - sin²(B)) / sin(B)
The sin(B) term in the numerator cancels out with the sin(B) in the denominator:
sin(B) + (1 - sin²(B)) / sin(B)
Now, let's simplify the second term in parentheses by dividing each term by sin(B):
sin(B) + (1 / sin(B)) - (sin²(B) / sin(B))
1 / sin(B) can be simplified as csc(B) using the reciprocal identity:
sin(B) + csc(B) - (sin(B) / sin(B))
The sin(B) term in the second term of parentheses cancels out:
sin(B) + csc(B) - 1
Finally, we have:
sin(B) + cos(B) cot(B) = csc(B) - 1
So, sin(B) + cos(B) cot(B) is not equal to csc(B), but is equal to csc(B) - 1.
Therefore, the initial statement sin(B) + cos(B) cot(B) = csc(B) is false.