Question

Albert wants to show that tan(theta)sin(theta)+cos(theta)=sec(theta). He writes the following proof:

Answer 1

Hello,

Answer B

He should write cos(theta)=cos²(theta)/cos(theta) to find a common denominator.

Answer 2

Albert must continue the following way to continue with the proof after the line $\frac{sin^2(\theta)}{cos(\theta)}+cos(\theta)$:

$\frac{sin^2(\theta)}{cos(\theta)}+cos(\theta)=\frac{sin^2(\theta)}{cos(\theta)}+cos(\theta)\times \frac{cos(\theta)}{sin(\theta)}$

=$\frac{sin^2(\theta)}{cos(\theta)}+\frac{cos^2(\theta)}{cos(\theta)}$

=$\frac{{sin^2(\theta)}+cos^2(\theta)}{cos(\theta)}$

=$\frac{1}{cos(\theta)}$

$=sec(\theta)$ which is the required answer.

Thus, as we can see, after the given line in the question, Albert should write $cos(\theta)=cos^2(\theta)/cos(\theta)$ or $\frac{cos^2(theta)}{cos(theta)}$ to find a common denominator.

Thus, Option B is the correct answer.