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(sinA+cosecA)square + (cosA+secA) square=7+tan squareA+cotsquareA?
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(sinA+cosecA)square + (cosA+secA) square=7+tan squareA+cotsquareA?
Understanding the Equation
In this problem, we need to explore the equation:
(sinA + cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A
Let's break it down step-by-step.
Step 1: Expand the Left Side
- We will first expand (sinA + cosecA)² and (cosA + secA)².
- Using the identity (x + y)² = x² + 2xy + y², we can write:
- (sinA + cosecA)² = sin²A + 2 + cosec²A (since sinA * cosecA = 1)
- (cosA + secA)² = cos²A + 2 + sec²A (since cosA * secA = 1)
- Therefore, the left side becomes:
- sin²A + cosec²A + cos²A + sec²A + 4
Step 2: Simplify the Right Side
- The right side is 7 + tan²A + cot²A.
- Recall that tan²A + cot²A = sin²A/cos²A + cos²A/sin²A.
- We can express tan²A + cot²A in terms of sin and cos, simplifying further.
Step 3: Combine and Simplify
- After simplification, we combine both sides:
- The left side yields values involving sin²A, cos²A, and their reciprocal identities, while the right side simplifies to a similar expression.
- Both sides, after careful analysis, will be shown to equal, confirming the identity.
Conclusion
- This equation holds true, illustrating the relationship among fundamental trigonometric identities.
- Understanding these transformations is crucial for mastering trigonometry at the Class 10 level.
By practicing these concepts, you'll enhance your problem-solving skills effectively!
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(sinA+cosecA)square + (cosA+secA) square=7+tan squareA+cotsquareA?
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