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**Ques: 1** Prove
that

**First Solution:**
We will show that

Indeed, by the **addition and subtraction
formulas**, we obtain

**Second Solution:**
Note that by the addition and subtraction formula,
we have

Hence and so

that is, , as desired.

%{font-family:verdana;font-size:17px;color:#150577}*Ques: 2* In
triangle ABC, show that

%

**Solution:** By the **extended law of sines**, we have

Applying the **double-angle formulas** and
**sum-to-product formulas** in the above relation
gives

by noting that because

**Note:** By symmetry, we have analogous formulas

and

%{font-family:verdana;font-size:17px;color:#150577}*Ques: 3* Let
for Prove
that for all real numbers x.%

**Solution:** We need to show that

for all real numbers x. Indeed, the left-hand side is equal to

**Ques: 4** A circle of
radius 1 is randomly placed in a 15Ã-36 rectangleABCD so that the
circle lies completely within the rectangle. Compute the
probability that the circle will not touch diagonal AC.

**Note :** In order for the
circle to lie completely within the rectangle, the center of the
circle must lie in a rectangle that is (15-2)x(36-2), or 13×34. The
requested probability is equal to the probability that the distance
from the circle's center to the diagonalAC is greater than 1, which
equals the probability that the distance from a randomly selected
point in the 13 × 34 rectangle to each side of triangles
ABC and CDA is
greater than 1. Let |AB| = 36 and |BC| = 15 (and so |AC| = 39).
Drawthree segments that are 1 unit away from each side of triangle
ABC and whose endpoints are on the sides.
Let E,F, and Gbe the three points of intersection nearest to A,B,
and C, respectively, of the three segments. Because the
corresponding sides of triangle ABC and
EFG are parallel, the two triangles are
similar to each other. The desired probability is equal to

|<>{border-color:white}. |<>{border-color:white;font-family:verdana}.
Because E is equidistant from sides AB and AC, E lies on the
bisector of Similarly, F and G lie on the bisectors of and
respectively. Hence lines AE, BF and CG meet I , the in center of
triangle ABC.|

**First
Solution:** Let and be the feet of the
perpendiculars from E and F to segment AB, respectively. Then
It is not difficult to see that Set Then and By either the **double-angle
formulas** or the **half-angle formulas,**

or

and we obtain It follows that or
Consequently, Hence
and

**Second
Solution:** Set and
Because E lies on the angle bisector of has the same slope as
that is, the slope of line AE is
Consequently, and the rest of the solution proceeds like that of the
first solution.

**Third
Solution:** Because the
corresponding sides of triangles ABC and
EFG are parallel, it follows that I is
also the incenter of triangle EFG and
that the triangles are **homothetic** (with I as the
center). If r is the inradius of triangle ABC, then r âˆ' 1 is the inradius of triangle
EFG; that is, the ratio of the similarity
between triangles EFG and ABC is
Hence the desired probability is

Note that

Solving the last equation gives r = 6, and so

**Ques: 5** Prove that
for all
where k is in Z.

**Solution:**
The equality is equivalent to

or

That is, which is evident.

**Note:** More generally, if are real
numbers different from where k is in Z, such that
then the relation

holds.

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