Inverse Trigonometric Functions

Let x=\tan\theta\) with . Use : Since , returns :

. Apply inverse on both sides (with principal ranges): .

Use the tangent addition formula . Take . Then . Apply : .

. must return an angle in . In that range, the angle whose sine is is , not .

returns an angle whose sine is the input. Thus satisfies . Applying undoes :

Let with . Then Since lies in and under usual exam assumptions , we write .

. Hence

With the standard principal ranges used in school exams, . Therefore

Using ,

Step 1: Recall that corresponds to in the principal range . Step 2: Thus, ---

Step 1: Recall the complementary angle identity between inverse cosine and inverse sine: Step 2: Rearranging gives: ---

Step 1: Using the odd function property of : for all . ---

Let We need to show: Using the addition formula for sine: Calculate : Then, Therefore, Now, Use sine addition formula again: Calculate and : Then, Thus,
Final answer:

Let: Recall the formula for sum of inverse cotangents: Calculate: We know: Therefore:

Let: Then, Using the tangent addition formula: Since and , it follows that: provided the expression is defined (i.e., ).

Step 1: Let and . Thus, We need to evaluate: Step 2: Use the tangent addition formula: Substitute the values: Step 3: Hence,

Step 1: Let . This means . Step 2: Use the identity , which leads to: Thus, Step 3: Therefore, ---

Step 1: Recall the principal value range of , which is . The cosine function has the property that for in the domain of the inverse cosine, when lies within the range . Step 2: Since lies within the range , we directly get: ---

Step 1: Recall that corresponds to an angle such that and . From the unit circle, we know: Step 2: Now evaluate: Step 3: Simplify the expression: Thus, the answer is . ---

Step 1: First, find the value of . This represents the angle such that , and . So, This gives an angle approximately equal to . Step 2: Now find the value of . Since the secant function is the reciprocal of the cosine function, we have , which implies . For , the angle lies in the range , and . Therefore, Step 3: Now calculate the difference: Thus, the answer is . ---

We are asked to find: We know that: Now, for , recall that: So, Now, subtracting the two terms: Thus, the correct answer is .

We are asked to find: We know that: For , since , we have: Thus, the expression becomes: Now simplify: Thus, the correct answer is .

We are given . We can use the formula for the sum of inverse tangents: So: First, simplify the numerator: And the denominator: So the expression becomes: Thus, the correct answer is .

We are given the equation: Since , the equation becomes: Now solve for : Taking the cosine of both sides: Thus, the correct value of is .

We are asked to differentiate , which is the inverse sine of . Step 1: Apply the derivative of inverse sine
The derivative of with respect to is: For , we apply the chain rule. Let , so:

We are asked to find: First, recall that: This is because the secant function is equal to 1 when the angle is . Thus, the expression simplifies to: Thus, the correct answer is .

We are given the expression . Recall that: So, Thus, the correct answer is .

We are asked to find: The principal range of is . Since is greater than , we need to find an equivalent angle in the range . The angle is greater than , so we subtract from it: Now, since is an even function, we know that: Therefore, we have: Thus, the correct answer is .

We are asked to find: We know that: Since the argument is negative, we are looking for the angle in the fourth quadrant. The value of is: Thus, the correct answer is .

We are given: Recognize that the derivative of is , which simplifies the integral. Therefore, we can rewrite the integral as: This is a standard integral whose result is . Thus, the correct answer is .

We are asked to differentiate . Step 1: Apply the derivative of inverse tangent
The derivative of is:
Step 2: Multiply by the constant 2
Since the expression is , we apply the constant multiple rule: