Question

Prove that sin²(θ) + cot(θ)tan(θ) = sec²(θ) - cos(θ).

A. sin²(θ) + cot(θ)tan(θ) = sec²(θ) - cos(θ)
B. sin²(θ) + cot(θ)tan(θ) = sec²(θ) + cos(θ)
C. sin²(θ) + cot(θ)tan(θ) = sec²(θ) + cos²(θ)
D. sin²(θ) + cot(θ)tan(θ) = sec²(θ)

Answer 1

The given equation can be proved using trigonometric identities. Starting with the left side of the equation, we simplify it step by step using identities such as sin²(θ) + cos²(θ) = 1 and sec²(θ) = 1 + tan²(θ). Finally, we obtain the right side of the equation, thus proving the given statement.

Step-by-step explanation:

To prove the given equation sin²(θ) + cot(θ)tan(θ) = sec²(θ) - cos(θ), we can use the trigonometric identity cot(θ) = cos(θ) / sin(θ). Let's start with the left side of the equation:

sin²(θ) + cot(θ)tan(θ)

= sin²(θ) + (cos(θ) / sin(θ)) * sin(θ)

= sin²(θ) + cos(θ)

Using the identity sin²(θ) + cos²(θ) = 1, we can rewrite the equation as:

= 1 - cos²(θ) + cos(θ)

= (1 + cos(θ))(1 - cos(θ)) + cos(θ)

= 1 - cos²(θ) + cos(θ) - cos²(θ)

= 1 - 2cos²(θ) + cos(θ)

Finally, using the identity sec²(θ) = 1 + tan²(θ), we can simplify the equation:

= 1 - 2(1 - sin²(θ)) + cos(θ)

= 1 - 2 + 2sin²(θ) + cos(θ)

= -1 + 2sin²(θ) + cos(θ)

= -cos²(θ) + sin²(θ) + cos(θ)

= sec²(θ) - cos(θ)