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Prove that sin square A by cos square A cos square A by sin squared equal to 1 by sin square A into cos square A minus 2 Pls anyone ans fastly.its time 2 go 2 schl.pls fstttt?
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Prove that sin square A by cos square A cos square A by sin squared ...
Proof of the Identity: sin²A/cos²A * cos²A/sin²A = 1/(sin²A*cos²A) - 2
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Proof using Trigonometric Identities:
1. Start with the left-hand side of the identity and simplify:
sin²A/cos²A * cos²A/sin²A = sin²A*sin²A / cos²A*cos²A
= (sin²A/cos²A)²
2. Use the Pythagorean Identity to rewrite sin²A/cos²A as tan²A + 1:
(sin²A/cos²A)² = (tan²A + 1)² / cos⁴A
= (tan⁴A + 2tan²A + 1) / cos⁴A
3. Use the Pythagorean Identity again to rewrite cos²A/sin²A as cot²A + 1:
(cos²A/sin²A)² = (cot²A + 1)² / sin⁴A
= (cot⁴A + 2cot²A + 1) / sin⁴A
4. Substitute these expressions back into the original identity:
(tan⁴A + 2tan²A + 1) / cos⁴A * (cot⁴A + 2cot²A + 1) / sin⁴A
= 1/(sin²A*cos²A) - 2
5. Simplify the expression on the right-hand side:
1/(sin²A*cos²A) - 2 = (1 - 2sin²A*cos²A) / (sin²A*cos²A)
= (1 - sin²2A) / (sin²A*cos²A)
= cos²2A / (sin²A*cos²A)
6. Substitute this expression back into the previous step:
(tan⁴A + 2tan²A + 1) / cos⁴A * (cot⁴A + 2cot²A + 1) / sin⁴A
= cos²2A / (sin²A*cos²A)
7. Simplify both sides of the equation:
tan⁴A + 2tan²A + 1 = cos²2A * sin⁴A / (cot⁴A + 2cot²A + 1)
8. Use the Pythagorean Identity to rewrite cot²A as 1 + tan²A:
tan⁴A + 2tan²A + 1 = cos²2A * sin⁴A / (1 + tan⁴A + 2tan²A)
9. Combine like terms:
tan⁴A - cos²2A * sin⁴A = -tan⁴A * cos²2A
10. Factor out tan²A:
tan²A (tan²A - cos²2A * sin²A) = -tan²A * cos²2A
11. Divide both sides by -tan²A:
tan²A - cos²2A * sin²A = -cos²2A
12. Use the double-angle identity for cosine:
tan²A - (1 - 2sin²A)sin²A = -1 + 2sin²A
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Proof using Trigonometric Identities:
1. Start with the left-hand side of the identity and simplify:
sin²A/cos²A * cos²A/sin²A = sin²A*sin²A / cos²A*cos²A
= (sin²A/cos²A)²
2. Use the Pythagorean Identity to rewrite sin²A/cos²A as tan²A + 1:
(sin²A/cos²A)² = (tan²A + 1)² / cos⁴A
= (tan⁴A + 2tan²A + 1) / cos⁴A
3. Use the Pythagorean Identity again to rewrite cos²A/sin²A as cot²A + 1:
(cos²A/sin²A)² = (cot²A + 1)² / sin⁴A
= (cot⁴A + 2cot²A + 1) / sin⁴A
4. Substitute these expressions back into the original identity:
(tan⁴A + 2tan²A + 1) / cos⁴A * (cot⁴A + 2cot²A + 1) / sin⁴A
= 1/(sin²A*cos²A) - 2
5. Simplify the expression on the right-hand side:
1/(sin²A*cos²A) - 2 = (1 - 2sin²A*cos²A) / (sin²A*cos²A)
= (1 - sin²2A) / (sin²A*cos²A)
= cos²2A / (sin²A*cos²A)
6. Substitute this expression back into the previous step:
(tan⁴A + 2tan²A + 1) / cos⁴A * (cot⁴A + 2cot²A + 1) / sin⁴A
= cos²2A / (sin²A*cos²A)
7. Simplify both sides of the equation:
tan⁴A + 2tan²A + 1 = cos²2A * sin⁴A / (cot⁴A + 2cot²A + 1)
8. Use the Pythagorean Identity to rewrite cot²A as 1 + tan²A:
tan⁴A + 2tan²A + 1 = cos²2A * sin⁴A / (1 + tan⁴A + 2tan²A)
9. Combine like terms:
tan⁴A - cos²2A * sin⁴A = -tan⁴A * cos²2A
10. Factor out tan²A:
tan²A (tan²A - cos²2A * sin²A) = -tan²A * cos²2A
11. Divide both sides by -tan²A:
tan²A - cos²2A * sin²A = -cos²2A
12. Use the double-angle identity for cosine:
tan²A - (1 - 2sin²A)sin²A = -1 + 2sin²A
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Prove that sin square A by cos square A cos square A by sin squared ...
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