arctan(cot10Pi/9)
x = arctan[cot(10pi/9)]
= arctan[1/tan(10pi/9)
= arctan[1/tan((9+1)pi/9)]
= arctan{1/tan([(9/9)+(1/9)]pi)}
= arctan[1/tan(pi/9)]
tan(x)-1/tan(pi/9)=0
[tan(pi/2)*tan(x)-1]/tan(pi/2)=0
tan(pi/2)*tan(x)-1=0
Express in sin and cos. Use identities also. (lol, go solve the details. too lazy to do it.)
--> cos[(pi/9)+x]=0
Find the arcos of both sides in radian form.
(pi/9)+x = pi/2
x = (pi/2)- (pi/9)
x = 7pi/18
arctan(cot(x))
= arctan(tan(pi/2-x))
= pi/2-x
This is not quite true when dealing with QIII angles...
To find the value of arctan(cot(10π/9)), we can follow these steps:
Step 1: Start with the expression cot(10π/9).
Step 2: cot(x) is the reciprocal of tan(x), so cot(10π/9) is equal to 1/tan(10π/9).
Step 3: tan(x) is defined as the ratio of sine(x) to cosine(x), so tan(10π/9) is equal to sin(10π/9) / cos(10π/9).
Step 4: Now, we have cot(10π/9) = 1 / (sin(10π/9) / cos(10π/9)).
Step 5: We can simplify this expression by multiplying the numerator and denominator by the reciprocal of sin(10π/9), which is csc(10π/9). This gives us cot(10π/9) = (1 * csc(10π/9)) / (sin(10π/9) * csc(10π/9) / cos(10π/9)).
Step 6: Simplifying further, cot(10π/9) = csc(10π/9) / (sin(10π/9) / cos(10π/9)).
Step 7: Using the identity csc(x) = 1/sin(x), we can rewrite this expression as cot(10π/9) = (1/sin(10π/9)) / (sin(10π/9) / cos(10π/9)).
Step 8: Cancel out the common term in the numerator and denominator: cot(10π/9) = 1 / cos(10π/9).
Step 9: Finally, we can calculate arctan(cot(10π/9)) by finding the angle whose cosine is equal to cos(10π/9).
To find the value of arctan(cot(10π/9)), we need to understand the relationships between trigonometric functions and inverse trigonometric functions.
First, let's simplify the expression cot(10π/9). The cotangent function is the reciprocal of the tangent function:
cot(x) = 1 / tan(x)
So, cot(10π/9) = 1 / tan(10π/9).
Next, recall that the tangent function is periodic with a period of π. Therefore, we can use the identity: tan(x + nπ) = tan(x), where n is an integer.
Now, let's simplify the expression tan(10π/9):
tan(10π/9) = tan(10π/9 + π) [Adding one period]
= tan((19π + 10π) / 9) [Combining the fractions]
= tan(29π / 9).
Since tan(x) has a period of π, we can simplify further using the identity: tan(x + nπ) = tan(x), where n is an integer.
tan(29π / 9) = tan(29π / 9 - 3π) [Subtracting three periods]
= tan((29π - 27π) / 9) [Combining the fractions]
= tan(2π / 9).
Now we have simplified the expression to tan(2π / 9). To find the value of arctan(tan(2π / 9)), we can use the fact that arctan(tan(x)) = x for -π/2 < x < π/2.
Therefore, arctan(tan(2π / 9)) = 2π / 9.
So, the value of arctan(cot(10π/9)) is 2π / 9.