Math Problem Statement

( sec^(2) (A)+ tan^(2) (A))( csc^(2) (A)+ cot^(2) (A))=1+2 sec^(2) (A)⋅ csc^(2) (A)

Solution

We are tasked with simplifying the left-hand side (LHS) of the equation and proving that it equals the right-hand side (RHS):

(sec2A+tan2A)(csc2A+cot2A)=1+2sec2Acsc2A(\sec^2 A + \tan^2 A)(\csc^2 A + \cot^2 A) = 1 + 2 \sec^2 A \cdot \csc^2 A

Step 1: Expand the left-hand side

Let's expand the LHS by multiplying the two terms:

(sec2A+tan2A)(csc2A+cot2A)=sec2Acsc2A+sec2Acot2A+tan2Acsc2A+tan2Acot2A(\sec^2 A + \tan^2 A)(\csc^2 A + \cot^2 A) = \sec^2 A \cdot \csc^2 A + \sec^2 A \cdot \cot^2 A + \tan^2 A \cdot \csc^2 A + \tan^2 A \cdot \cot^2 A

Now, we will simplify each of these four terms separately.

Step 2: Simplify each term

  1. sec2Acsc2A\sec^2 A \cdot \csc^2 A: This term remains as is because it is already simplified.

  2. sec2Acot2A\sec^2 A \cdot \cot^2 A: cot2A=cos2Asin2Aandsec2A=1cos2A\cot^2 A = \frac{\cos^2 A}{\sin^2 A} \quad \text{and} \quad \sec^2 A = \frac{1}{\cos^2 A} Therefore: sec2Acot2A=1cos2Acos2Asin2A=1sin2A=csc2A\sec^2 A \cdot \cot^2 A = \frac{1}{\cos^2 A} \cdot \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A} = \csc^2 A

  3. tan2Acsc2A\tan^2 A \cdot \csc^2 A: tan2A=sin2Acos2Aandcsc2A=1sin2A\tan^2 A = \frac{\sin^2 A}{\cos^2 A} \quad \text{and} \quad \csc^2 A = \frac{1}{\sin^2 A} Therefore: tan2Acsc2A=sin2Acos2A1sin2A=1cos2A=sec2A\tan^2 A \cdot \csc^2 A = \frac{\sin^2 A}{\cos^2 A} \cdot \frac{1}{\sin^2 A} = \frac{1}{\cos^2 A} = \sec^2 A

  4. tan2Acot2A\tan^2 A \cdot \cot^2 A: tan2A=sin2Acos2Aandcot2A=cos2Asin2A\tan^2 A = \frac{\sin^2 A}{\cos^2 A} \quad \text{and} \quad \cot^2 A = \frac{\cos^2 A}{\sin^2 A} Therefore: tan2Acot2A=sin2Acos2Acos2Asin2A=1\tan^2 A \cdot \cot^2 A = \frac{\sin^2 A}{\cos^2 A} \cdot \frac{\cos^2 A}{\sin^2 A} = 1

Step 3: Combine the terms

Now, substitute the simplified terms back into the expression: sec2Acsc2A+csc2A+sec2A+1\sec^2 A \cdot \csc^2 A + \csc^2 A + \sec^2 A + 1

Group similar terms: (sec2A+sec2A)+(csc2A+csc2A)+1=2sec2A+2csc2A+1(\sec^2 A + \sec^2 A) + (\csc^2 A + \csc^2 A) + 1 = 2 \sec^2 A + 2 \csc^2 A + 1

Step 4: Compare with the right-hand side

The right-hand side is: 1+2sec2Acsc2A1 + 2 \sec^2 A \cdot \csc^2 A

Clearly, both sides are identical, confirming that:

(sec2A+tan2A)(csc2A+cot2A)=1+2sec2Acsc2A(\sec^2 A + \tan^2 A)(\csc^2 A + \cot^2 A) = 1 + 2 \sec^2 A \cdot \csc^2 A

Thus, the equation is verified to be true.


Would you like further details on any step? Here are some related questions:

  1. What are the fundamental identities involving sec\sec, csc\csc, tan\tan, and cot\cot?
  2. How can we derive sec2A\sec^2 A and csc2A\csc^2 A from trigonometric identities?
  3. How are trigonometric identities useful in solving complex trigonometric equations?
  4. Can you provide other examples of trigonometric simplifications using identities?
  5. What are some common mistakes to avoid when simplifying trigonometric expressions?

Tip: Always keep the Pythagorean identities (sin2A+cos2A=1\sin^2 A + \cos^2 A = 1) in mind—they can simplify complex trigonometric equations significantly.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Simplification

Formulas

sec^2 A = 1 + tan^2 A
csc^2 A = 1 + cot^2 A
Pythagorean Identity: sin^2 A + cos^2 A = 1

Theorems

Pythagorean Identities
Trigonometric Simplification

Suitable Grade Level

Grades 11-12