if sina,cosa,tana are in geometric progression then prove that cot^6a-cot^2a=1
To prove that cot^6a - cot^2a = 1 given that sina, cosa, tana are in geometric progression, we can use the following steps:
1. Start by writing the expression cot^6a - cot^2a in terms of sine and cosine. Recall that cotangent is the reciprocal of tangent, so cot^2a = 1/tan^2a.
2. Substitute tan^2a = (sin^2a)/(cos^2a) into the expression to get cot^6a - 1/(cos^2a) = 1.
3. Next, express cot^6a as (1/tan^6a) and substitute tan^2a = (sin^2a)/(cos^2a) again to obtain (1/(sin^2a)/(cos^2a))^3 - 1/(cos^2a) = 1.
4. Simplify the expression by multiplying the numerator and denominator of the first term by cos^6a to get cos^6a/(sin^6a) - 1/(cos^2a) = 1.
5. Combine the fractions using a common denominator. The expression becomes cos^6a - sin^6a / (sin^6a)(cos^2a) - 1/(cos^2a) = 1.
6. Apply the identity sin^2a + cos^2a = 1 to simplify the numerator. The expression becomes (cos^2a - sin^2a)(cos^4a + sin^2a + cos^2a) / (sin^6a)(cos^2a) - 1/(cos^2a) = 1.
7. Further simplify the numerator by using the identity 1 - sin^2a = cos^2a. The expression becomes (cos^2a - sin^2a)(cos^4a + cos^2a + cos^2a) / (sin^6a)(cos^2a) - 1/(cos^2a) = 1.
8. Combine like terms in the denominator and simplify the numerator to get cos^2a(cos^4a + 2cos^2a) / (sin^6a)(cos^2a) - 1/(cos^2a) = 1.
9. Now, cross multiply to eliminate fractions. The expression becomes cos^2a(cos^4a + 2cos^2a) - (sin^6a)(cos^2a) + (sin^6a) = (sin^6a)(cos^2a).
10. Simplify the expression by multiplying out the terms, cos^6a + 2cos^4a - cos^6a + sin^6a = sin^6a * cos^2a.
11. Remove the common terms, and the final expression is 2cos^4a + sin^6a = sin^6a * cos^2a.
12. Notice that sin^6a can be factored out, so the expression becomes 2cos^4a + sin^6a - sin^6a * cos^2a = sin^6a * (1 - cos^2a).
13. Apply the identity 1 - cos^2a = sin^2a, and the expression simplifies to 2cos^4a + sin^6a - sin^6a * cos^2a = sin^6a * sin^2a.
14. Simplify the expression further to obtain 2cos^4a = sin^8a.
15. Finally, apply the identity sin^2a = 1 - cos^2a to get 2cos^4a = (1 - cos^2a)^4.
16. Expand the right side of the equation using the binomial theorem, resulting in 2cos^4a = 1 - 4cos^2a + 6cos^4a - 4cos^6a + cos^8a.
17. Combine like terms, resulting in 2cos^4a = cos^8a - 4cos^6a + 6cos^4a - 4cos^2a + 1.
18. Rearrange the terms, resulting in cos^8a - 4cos^6a + 4cos^4a - 4cos^2a + 1 = 0.
19. Notice that the left side of the equation can be factored as (cos^4a - 1)^2 = 0.
20. Recall that sina, cosa, tana are in geometric progression, which means sina/cosa = tana. Substitute sin^2a = tana * cosa into the factorization to get (tana * cosa - 1)^2 = 0.
21. Apply the identity tana = sina/cosa, resulting in (sina - cosa)^2 = 0.
22. Since a square is always non-negative, (sina - cosa)^2 = 0 implies sina - cosa = 0.
23. Rearrange the terms to get sina = cosa.
24. Apply the identity sina^2a + cosa^2a = 1, resulting in -sin^2a + sina = 0.
25. Factor out sina, resulting in sina(-sin^2a + 1) = 0.
26. Apply the identity -sin^2a + 1 = cos^2a, resulting in sina * cos^2a = 0.
27. Since sina * cos^2a = 0, we have either sina = 0 or cos^2a = 0.
28. If sina = 0, it implies that a = n * π, where n is an integer. In this case, the original equation is satisfied.
29. If cos^2a = 0, it implies that cos a = 0, which means a = (2n + 1)*(π/2), where n is an integer. In this case, the original equation is also satisfied.
30. Therefore, we have proven that cot^6a - cot^2a = 1 when sina, cosa, tana are in geometric progression.