Inverse Trigonometric Functions

sin^-1x + sin^-1y + sin^-1z =
We know that the principal values of sin^-1θ lie between - and . Since the sum of the three angles is equal to , it means that each angle must be equal to . Therefore, we have:
sin^-1x =
sin^-1y =
sin^-1z =
Taking the sine of both sides of these equations:
sin(sin^-1x) = sin
sin(sin^-1y) = sin
sin(sin^-1z) = sin
Using the inverse sine function's property sin(sin^-1θ) = θ, we simplify:
x = 1
y = 1
z = 1
Now, we can calculate the sum of their 100^th powers:
x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ = 1¹⁰⁰ + 1¹⁰⁰ + 1¹⁰⁰ = 1 + 1 + 1 = 3
Therefore, the value of x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ is 3.
Among the given options, (C) 3 is the correct answer.

Given Equation:
The given equation is: We can simplify this equation using trigonometric identities.

Step 1: Use the Sum-to-Product Identity:
Apply the sum-to-product identity for sine functions: Using this identity for , we get: So, the equation becomes:

Step 2: Factor the Equation:
We can factor out from both sides of the equation: This equation is satisfied when either or .

Step 3: Solve for :
For , the solutions occur when: This gives: So, the possible solutions for are

Step 4: Solve for :
For , solve for : The solutions to occur when: So, we have: Therefore, the solutions for are

Final Answer:
Combining the solutions from both cases, the possible solutions to the equation are: Hence, the solutions are .

We are given the equation:

Step 1: Rearrange the equation: We can rearrange the equation as:

Step 2: Simplify further: Simplifying the above equation gives:

Step 3: Simplify the terms: This equation can be rewritten as:

Step 4: Deduce the solution: From the equation, we can deduce that:

Final Step: Solve for : Therefore, the solutions for are:

Given Equation:
We are given the equation:

Step 1: Use the Double Angle Identity for Tangent:
Recall that the identity for the tangent of a half-angle is: To apply this identity, let , so that: Now, the left-hand side is , and we can equate the two sides:

Step 2: Make a Substitution:
Let , so that . Now the equation becomes:

Step 3: Simplify and Solve for :
Recognize that the left-hand side simplifies using the identity for tangent subtraction: Thus, the equation becomes: Solving this equation by trial or approximation, we find that:

Final Answer:
Therefore, the solution for is:

Let's find the principal value of .

• is in the second quadrant of the unit circle.
• In the second quadrant, the sine function is positive.
• Using the identity , we have:
• . Hence:

• .
• lies within the principal value range .