Math Problem Statement

show that underoot(sec^2(x) -1) + underoot(cosec^2(x)-1) = sec(x)cosec(x)

Solution

To prove that:

sec2(x)1+cosec2(x)1=sec(x)cosec(x)\sqrt{\sec^2(x) - 1} + \sqrt{\cosec^2(x) - 1} = \sec(x)\cosec(x)

Step 1: Simplifying each term

Start by recalling the following trigonometric identities:

sec2(x)1=tan2(x)\sec^2(x) - 1 = \tan^2(x) cosec2(x)1=cot2(x)\cosec^2(x) - 1 = \cot^2(x)

So, the left-hand side becomes:

sec2(x)1=tan2(x)=tan(x)=tan(x)(since tan(x)0 in the given domain)\sqrt{\sec^2(x) - 1} = \sqrt{\tan^2(x)} = |\tan(x)| = \tan(x) \quad (\text{since } \tan(x) \geq 0 \text{ in the given domain}) cosec2(x)1=cot2(x)=cot(x)=cot(x)(since cot(x)0 in the given domain)\sqrt{\cosec^2(x) - 1} = \sqrt{\cot^2(x)} = |\cot(x)| = \cot(x) \quad (\text{since } \cot(x) \geq 0 \text{ in the given domain})

Thus, the left-hand side becomes:

tan(x)+cot(x)\tan(x) + \cot(x)

Step 2: Express the Right-Hand Side

The right-hand side of the equation is:

sec(x)cosec(x)\sec(x)\cosec(x)

Now, express sec(x)\sec(x) and cosec(x)\cosec(x) in terms of sine and cosine:

sec(x)=1cos(x)andcosec(x)=1sin(x)\sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \cosec(x) = \frac{1}{\sin(x)}

Thus, the right-hand side becomes:

sec(x)cosec(x)=1cos(x)1sin(x)=1cos(x)sin(x)\sec(x)\cosec(x) = \frac{1}{\cos(x)} \cdot \frac{1}{\sin(x)} = \frac{1}{\cos(x)\sin(x)}

Step 3: Equating the Two Sides

Now, we need to show that:

tan(x)+cot(x)=1cos(x)sin(x)\tan(x) + \cot(x) = \frac{1}{\cos(x)\sin(x)}

We know that:

tan(x)=sin(x)cos(x)andcot(x)=cos(x)sin(x)\tan(x) = \frac{\sin(x)}{\cos(x)} \quad \text{and} \quad \cot(x) = \frac{\cos(x)}{\sin(x)}

Thus:

tan(x)+cot(x)=sin(x)cos(x)+cos(x)sin(x)\tan(x) + \cot(x) = \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}

Take the common denominator:

tan(x)+cot(x)=sin2(x)+cos2(x)sin(x)cos(x)\tan(x) + \cot(x) = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)}

Using the Pythagorean identity sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1, this simplifies to:

tan(x)+cot(x)=1sin(x)cos(x)\tan(x) + \cot(x) = \frac{1}{\sin(x)\cos(x)}

This is exactly the same as the right-hand side of the equation.

Conclusion:

Thus, we have shown that:

sec2(x)1+cosec2(x)1=sec(x)cosec(x)\sqrt{\sec^2(x) - 1} + \sqrt{\cosec^2(x) - 1} = \sec(x)\cosec(x)

Do you have any questions or would you like more details on any specific part of this explanation?

Relative Questions:

  1. What are the basic trigonometric identities?
  2. How do you derive the secant and cosecant functions from sine and cosine?
  3. What is the Pythagorean identity in trigonometry?
  4. Can the tangent and cotangent be expressed in other ways using sine and cosine?
  5. How do we handle absolute values in trigonometric identities?
  6. What are some other common trigonometric identities used in calculus?
  7. How can you prove the identity sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1?
  8. What happens if you change the domain for tan(x)\tan(x) and cot(x)\cot(x)?

Tip:

Always break down complex trigonometric identities by simplifying each component step-by-step using known identities.