Trigonometry

Step 1: Simplify the expression for .
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Step 2: Simplify the expression for .
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Step 3: Simplify the expression for .
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Step 4: Compare the values of , , and .
, , .
Step 5: Arrange , , and in ascending order.
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Step 6: Match the order with the given options.
The ascending order is R, Q, P.

Step 1: Find and . Given , we calculate and using the identity .

Step 2: Find and . Given , we calculate using the identity .

Step 3: Apply the angle addition formula for . We use the identity , and the double angle formulas for sine and cosine to calculate the value of .

Step 4: Final result. The final result is:

A. ΣcotA

This represents the sum of the cotangents of the angles of a triangle. It is equivalent to (a² + b² + c²) / 4Δ where a, b, and c are the side lengths and Δ is the triangle's area.
Therefore, A matches with II.

B. Σcot(A/2)

This represents the sum of the cotangents of the half-angles. This equals (a + b + c)² / 4Δ.
Thus, B matches with I.

C. tanA : tanB : tanC = 1:2:3

The ratio of sines of the angles is proportional to the sides opposite to the respective angles. So, sin A : sin B : sin C = √5 : 2√2 : 3.
C matches with V.

D. cot(A/2) : cot(B/2) : cot(C/2) = 3:7:9

The ratio of cotangents of half-angles is proportional to (s-a):(s-b):(s-c), where 's' is the semi-perimeter. Since a+b+c = 2s, the sides a:b:c are proportional to 12:5:13.
Therefore, D matches with III.

Step 1: Analyze the given equation.
We are given: $ : Since , we have: \) :
Let . Then:
\) : Let: \) depends on , which ranges from 0 to 1 as varies over its domain. The term also varies continuously with . Importantly, the equation involves trigonometric and inverse trigonometric functions, which can lead to infinitely many solutions due to periodicity and symmetry.
Step 4: Conclusion.
Given the nature of the functions involved, the equation has infinitely many solutions. This is because the combination of and allows for multiple values of that satisfy the equation.
Step 5: Final Answer.
\) $

Step 1: Use the half-angle formula for tangent. We are given . We need to find .
The half-angle formula for tangent is given by: Substitute the value of :
Step 2: Determine the sign of based on the quadrant of . We are given that . Since cosine is negative, must lie in either the second quadrant or the third quadrant. The problem states that does not lie in the second quadrant.
Therefore, must lie in the third quadrant. If is in the third quadrant, its general form can be written as for any integer .
Let's consider the principal range for : . Now, divide the inequality by 2 to find the range for : An angle lying between and is in the second quadrant.
Step 3: Conclude the value of . In the second quadrant, the tangent function is negative.
Since we found , and is in the second quadrant, we must choose the negative sign.
Therefore, . The final answer is -2

Step 1: Determine the quadrant of .
We are given two conditions:
1. (sine is negative)
2. (cosine is negative)
For both sine and cosine to be negative, the angle must lie in the third quadrant.
Step 2: Find the value of .
We can find using the given values of and :
Step 3: Write the general solution for .
Let . By definition of the inverse tangent function, is an angle in the first quadrant ( ) such that .
The general solution for is given by: Substituting : Step 4: Filter the general solution to ensure is in the third quadrant.
We know from Step 1 that must be in the third quadrant.
Let's test values of :
If is an even integer (e.g., for ):
This angle lies in the first quadrant (since is in the first quadrant and adding does not change the quadrant). In the first quadrant, sine and cosine are positive, which contradicts the given conditions. If is an odd integer (e.g., for ):
For , . Adding to a first-quadrant angle places it in the third quadrant. In the third quadrant, sine and cosine are both negative, which is consistent with the given conditions.
This form covers all angles in the third quadrant that have a tangent of .
Therefore, the general solution satisfying both conditions is , where .