Trigonometric Functions

Given,

Now,

Given,
LHS
Since,
But
Hence, no solution exist.

Given, and

Given,
Since, lies in the second quadrant.

We know that in


,
where is circumradius of .




is right angled.

In




In

Given




....(i)






[from E(i)]

Given,
If we draw the graphs of and , clearly
when

Given,
or
Now, if , then only one solution i.e. is possible.
Also, if , then two solutions are possible.
Hence, there are total 3 solutions.

We are given the function:
We need to determine which value of equals 1.
Step 1: Analyze the general behavior of the function for small values of .
and both have simple expansions for small , allowing us to evaluate their behavior for specific values of like .
Notice that for , the terms of the product are very close to 1 for the first few terms.
Step 2: Examine .
Calculating the exact value of reveals that it simplifies to 1. This happens because the initial terms of the product become balanced and also simplifies in a way that results in 1.
Step 3: Check other values of .
- For and , the values do not simplify to 1. Therefore, the correct answer is .

We are asked to evaluate the following limit:
Step 1: Break down the components inside the tangent.
Notice that the expression has both and . For small values of , we know that: So, for small , the expression simplifies to: Since is a constant, we now have a simple expression involving small . Evaluating it carefully reveals that the limit evaluates to .
Step 2: Simplify the limit.
Thus, the value of the limit is:
Final Answer: .