Properties Of Inverse Trigonometric Functions

Step 1: Concept
only if . Otherwise, we must find an angle in the principal range.
Step 2: Analysis
radians is approximately . We know . Thus, is near . is not correct because . .
Step 3: Conclusion
, which lies within . So, .
Final Answer: (C)

Step 1: Concept
Identify the trigonometric substitution. .
Step 2: Analysis
Put , then .
Step 3: Conclusion
Differentiating with respect to , .
Final Answer: (A)

Step 1: Concept
Use the identity if .
Step 2: Analysis
. Now, add : .
Step 3: Conclusion
Result .
Final Answer: (A)

Step 1: Concept
Use trigonometric substitution to simplify the inverse function.
Step 2: Analysis
Substitute . Then . The expression becomes .
Step 3: Conclusion
Since , the result is .
Final Answer: (A)

Step 1: Concept
Use the identity .
Step 2: Analysis
Let and .
Result
.
Step 3: Conclusion
.
Final Answer: (A)

Step 1: Concept
Use the identity .
Step 2: Analysis
Let . Then . Using . .
Step 3: Conclusion
.
Final Answer: (A)

Step 1: Principal values
. . . .
Step 2: Substitution
.
Step 3: Calculation
.
Final Answer: (D)

Step 1: Concept
only if . Otherwise, we must find an angle in the principal range.
Step 2: Analysis
radians is approximately . We know . Thus, is near . is not correct because . .
Step 3: Conclusion
, which lies within . So, .
Final Answer: (C)

Step 1: Understand the given equation.
We are given the equation:
where and are inverse trigonometric functions. We need to solve for in the interval .

Step 2: Simplify the inverse trigonometric functions.
Recall the identity for . Applying this to the given equation: we can reduce the equation to:

Step 3: Solve for .
Now, use the properties of inverse trigonometric functions to find possible values of . The simplifications will lead to values for within the interval .

Step 4: Find the number of solutions.
By solving the equation, we find there are exactly 2 solutions for in the interval .

Step 5: Conclusion.
Thus, the number of solutions to the equation is 2, corresponding to option (B).

Step 1: Recall the range of the inverse cosine function.
The range of the inverse cosine function is , meaning that the output of the inverse cosine is always within this interval.

Step 2: Analyze the angle .
The angle lies in the third quadrant. To find the corresponding angle in the range , we need to use the reference angle.

Step 3: Find the reference angle.
The reference angle for is:

Step 4: Apply the inverse cosine.
Since , and the inverse cosine function returns values in the range , we have:

Step 5: Conclusion.
Thus, the value of is , corresponding to option (C).

Step 1: Recall the domain and range of .
The inverse sine function is defined for values of such that . Hence, the domain of is .

Step 2: Find the range of .
The range of the inverse sine function is: This is because the range of the sine function is .

Step 3: Conclusion.
Thus, the domain of is , and the range is . Therefore, the correct answer is option (B).

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}


As ,
(since )
Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
Let

Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}





LHS , RHS (since )
No solution.
Step 3: Final Answer

Concept: Use half-angle identities to simplify the expression inside the inverse cotangent.

Step 1: Express in terms of half-angles. For (appropriate domain), and . So:

Step 2: Simplify the fraction. Numerator: Denominator: Thus:

Step 3: Apply inverse cotangent. (provided is in the principal range of , which is ).

Concept:

Step 1: Let .


Step 2: Use .
Conclusion:

Step 1: Evaluate .


Step 2: Substitute.


Step 3: Convert to sine.
Let

Concept: where , , .

Step 1: Use given condition.

Step 2: Apply identity.

Step 3: Condition for fraction = 0. A fraction is zero when numerator = 0 and denominator :

Concept: Let angles:

Step 1: Find cos A.


Step 2: Find sin B.


Step 3: Use identity.
Hence required value:

Concept: If both numbers and their inverse tangent are in AP only possible when all are equal.

Step 1: Use AP condition.


Step 2: Apply for tan .


Step 3: Conclusion.
Only possible when:

Concept: Use standard identities

Step 1:Evaluate Let , so

Step 2:Evaluate Let , so Using triangle: Half-angle formula:

Step 3:Verify relation