Trigonometric Functions

We are given a right-angled triangle where . Using the properties of the incenter and applying the standard formula for the incircles, we simplify the equation and obtain the value as 2.

Step 1: Compute the integral .
Use the Weierstrass substitution: let , so: Also, , since , so , and . Thus: Substitute back :
Step 2: Compare with the given form .
We need: Rewrite the left-hand side: Multiply numerator and denominator by : This form is complex to match directly with . Instead, differentiate both sides to compare:
Right-hand side: Left-hand side (from our integral): This comparison is complex, so test with a specific . Set : This doesn’t match the options. Recompute the integral correctly: The standard result for is indeed , so: This matches option (3). Final Answer:

We are given the trigonometric equations involving and . By solving these equations and using the identity for , we find that the value is 4.

We are given the equation in a triangle. By using standard trigonometric identities and simplifying, we find the value of the expression to be .

We are given the equation . By simplifying the terms using trigonometric identities, we find that the expression simplifies to .

Given that are in arithmetic progression, and , we can use Heron's formula to find the area of the triangle. By substituting the given values and solving the equation, we find the area of to be square centimeters.

The given function is:
We need to find the value of . Let's begin by simplifying the function. Notice that contains trigonometric terms. We will attempt to evaluate it at . The key observation here is that the numerator and denominator share similar forms, and since trigonometric functions like sine and cosine repeat periodically, the values of and will follow the same periodic pattern. For , since the periodic nature of sine and cosine ensures that both and have values that result in the overall simplification of the expression, we can evaluate the expression as: Thus, the value of is .

The function given is . To find the number of turning points in the interval , we begin by finding its derivative.

The first derivative is:

Using the double angle identity , we rewrite the equation:

Simplifying:

Setting for critical points,

Using , substitute:

Let , then:

Using the quadratic formula, with , , ,

Therefore,

Turning Points are at .
For , .
For , within .

Thus, there are 3 turning points: .

The correct answer is 3.

Step 1: Use identity for .
Step 2: Use triangle to find and .
Given: .
Let opposite = 3, adjacent = 4. Then hypotenuse: Step 3: Plug values into identity.

Given in the first quadrant, where all trigonometric functions are positive. Let the opposite side be 4 and the adjacent side be 3 in a right triangle. The hypotenuse is: Verify: Thus, the value of is: