Thank you so much in advance!
Verify each identity. Work one side (the harder side) Please show me the work
sin θ +cos θ cot θ= csc θ
Cos^2 θ + 1 =2cos^2 θ +sin^2 θ
sin T + cos T cot T
sin T + cos T cos T/sin T
(sin^2 T + cos^2 T ) /sin T
1/sin T
which is csc T
=============================
LOL, subtract cos^2 T from both sides
1 = cos^2 T + sin^2 T
what else is new :)
To verify the given identities, we need to manipulate one side of the equation to make it equal to the other side. Let's start with the first identity:
sin θ + cos θ cot θ = csc θ
To work on the harder side, let's simplify the right-hand side (RHS) of the equation.
RHS = csc θ
Using the reciprocal identity, csc θ = 1/sin θ, we can rewrite the RHS:
RHS = 1/sin θ
Now, let's focus on the left-hand side (LHS) of the equation and manipulate it to match with the RHS.
LHS = sin θ + cos θ cot θ
We can rewrite cot θ as cos θ/sin θ using the reciprocal identity, cot θ = cos θ/sin θ:
LHS = sin θ + cos θ * (cos θ/sin θ)
Now, let's simplify the expression on the LHS:
LHS = sin θ + (cos^2 θ / sin θ)
Next, we need to find the common denominator between sin θ and cos^2 θ. The common denominator is sin θ, so let's multiply the first term by cos^2 θ / cos^2 θ:
LHS = (sin θ * cos^2 θ / cos^2 θ) + (cos^2 θ / sin θ)
Now, let's simplify further:
LHS = (sin θ * cos^2 θ + cos^2 θ) / cos^2 θ * sin θ
Now, factor out cos^2 θ from the numerator:
LHS = (cos^2 θ * (sin θ + 1)) / cos^2 θ * sin θ
Finally, cancel out the cos^2 θ terms in the numerator and denominator:
LHS = (sin θ + 1) / sin θ
Now, we can see that the LHS matches the RHS, which means the identity is valid.
sin θ + cos θ cot θ = csc θ is verified.
Let's move on to the second identity:
cos^2 θ + 1 = 2cos^2 θ + sin^2 θ
We'll work on the harder side, which is the left-hand side (LHS):
LHS = cos^2 θ + 1
To simplify this expression, let's subtract 2cos^2 θ from both sides:
LHS - 2cos^2 θ = cos^2 θ + 1 - 2cos^2 θ
Simplifying further:
LHS - 2cos^2 θ = 1 - cos^2 θ
Now, let's rewrite the right-hand side (RHS) using the trigonometric identity, sin^2 θ + cos^2 θ = 1:
RHS = sin^2 θ + cos^2 θ = 1
Substituting back into the equation:
LHS - 2cos^2 θ = RHS
cos^2 θ + 1 - 2cos^2 θ = 1
Next, let's combine like terms on the left-hand side:
-cos^2 θ - cos^2 θ + 1 = 1
Simplifying further:
-2cos^2 θ + 1 = 1
Now, subtract 1 from both sides:
-2cos^2 θ = 0
Next, divide both sides by -2 to solve for cos^2 θ:
cos^2 θ = 0
Taking the square root of both sides:
cos θ = ±√0
Since the square root of 0 is 0, we have two solutions:
cos θ = 0
Now, we can use these solutions to verify the identity. Plugging in cos θ = 0 into both sides of the equation:
LHS = cos^2 θ + 1
LHS = 0^2 + 1
LHS = 0 + 1
LHS = 1
RHS = 2cos^2 θ + sin^2 θ
RHS = 2(0^2) + sin^2 θ
RHS = 2(0) + sin^2 θ
RHS = 0 + sin^2 θ
RHS = sin^2 θ
As the LHS equals 1 and the RHS equals sin^2 θ, it is clear that the identity is not valid for all values of θ. The identity only holds true when cos θ = 0.
I hope this explanation helps you verify the given identities!