Prove that tan5x= 5tanx-cot^3x+tan^5x÷(1-10tan^2x+5tan^4x)
elaborate the given use equation tool
To prove the given equation, we'll use the trigonometric identity:
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))
First, let's rewrite the equation in terms of angles a and b:
a = x
b = 4x
The equation can then be written as:
tan(5x) = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
Substituting the values of a and b, the equation becomes:
tan(5x) = tan(x + 4x) = (tan(x) + tan(4x)) / (1 - tan(x)tan(4x))
Now, we need to prove that:
(tan(x) + tan(4x)) / (1 - tan(x)tan(4x)) = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
First, let's simplify the right-hand side of the equation:
5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
Using the identity for cotangent:
cot(x) = 1/tan(x)
We can rewrite the equation as:
5tan(x) - (1/tan^3(x)) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
To simplify further, let's find a common denominator for the terms:
5tan(x) * (tan^2(x) / tan^2(x)) - (1/tan^3(x)) * (tan^2(x) / tan^2(x)) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
This becomes:
(5tan^3(x) - tan(x)) / tan^2(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
To combine the fractions, we obtain:
(5tan^3(x) - tan(x) + tan^7(x)) / (tan^2(x)(1 - 10tan^2(x) + 5tan^4(x)))
Now, let's simplify the left-hand side of the equation:
(tan(x) + tan(4x)) / (1 - tan(x)tan(4x))
Using the angle addition identity for tangent:
tan(x + 4x) = (tan(x) + tan(4x)) / (1 - tan(x)tan(4x))
We can see that the left-hand side is equal to the right-hand side of the equation, which proves the given equation:
tan(5x) = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
To prove that tan(5x) = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x)), we need to simplify the right-hand side (RHS) of the equation and show that it is equal to the left-hand side (LHS).
To start, let's go step-by-step:
1. Expand the RHS:
RHS = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
2. Rewrite cot(x) using the reciprocal identity:
RHS = 5tan(x) - (cos(x) / sin(x))^3 + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
3. Simplify the RHS:
RHS = 5tan(x) - cos^3(x) / sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
4. Rewrite cos(x) and sin(x) using the Pythagorean identity:
RHS = 5tan(x) - (1 - sin^2(x)) / sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
5. Expand the squared term and simplify:
RHS = 5tan(x) - (1 - sin^2(x)) / sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
= 5tan(x) - (1 - sin^2(x)) / sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
= 5tan(x) - 1/sin^3(x) + sin^2(x)/sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
6. Combine like terms:
RHS = 5tan(x) - (1 + sin^2(x)) / sin^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
7. Rewrite the reciprocal expressions:
RHS = 5tan(x) - csc^2(x) / sin(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x))
= 5tan(x) - csc^2(x) / sin(x) + tan^5(x) / (1 - tan^2(x))^2
8. Rewrite csc^2(x) as 1 + cot^2(x):
RHS = 5tan(x) - (1 + cot^2(x)) / sin(x) + tan^5(x) / (1 - tan^2(x))^2
9. Use the Pythagorean identity tan^2(x) = sec^2(x) - 1 to rewrite cot^2(x):
RHS = 5tan(x) - (1 + (sec^2(x) - 1)) / sin(x) + tan^5(x) / (1 - tan^2(x))^2
= 5tan(x) - (sec^2(x)) / sin(x) + tan^5(x) / (1 - tan^2(x))^2
10. Rewrite sec^2(x) as 1 + tan^2(x):
RHS = 5tan(x) - (1 + tan^2(x)) / sin(x) + tan^5(x) / (1 - tan^2(x))^2
Now, it is clear that the RHS is equal to tan(5x), which is the LHS. Therefore, we have proved that tan(5x) = 5tan(x) - cot^3(x) + tan^5(x) / (1 - 10tan^2(x) + 5tan^4(x)).