If sec(x+y),sec(x),sec(x-y) are in A.P. then prove that cosx= +-√2 cos y/2 where cosx and cosy are not equals to 1,

because the difference of an AP is constant,

sec(x) - sec(x+y) = sec(x-y) - sec(x)

or, converting to cosines for ease of calculation,

cos(x+y) cos(x-y) - cos(x) cos(x-y) = cos(x) cos(x+y) - cos(x+y) cos(x-y)

2(cosx cosy - sinx siny)(cosx cosy + sinx siny) = cosx(cosx cosy + sinx siny) + cosx(cosx cosy - sinx siny)

cos^2(x)cos^2(y) - sin^2(x)sin^2(y) = cos^2(x)cosy

cos^2(x)cos^2(y) - (1-cos^2(x)-cos^2(y)+cos^2(x)cos^2(y)) = cos^2(x)cos(y)

cos^2(x)(cos(y)-1) = cos^2(y)-1
cos^2(x) = cos(y)+1
cos^2(x) = 2cos^2(y/2)
cos(x) = ±√2 cos(y/2)

Well, well, let's have some fun with this one! If sec(x+y), sec(x), and sec(x-y) are in arithmetic progression (A.P.), then we find ourselves in an interesting mathematical circus. Let's see if we can clown around a bit and prove that cos(x) = ±√2 cos(y/2), with the condition that cos(x) and cos(y) are not equal to 1.

First, let's recall some trigonometric clown tricks:

1. Secant (sec(x)) is the reciprocal of cosine (cos(x)). So, sec(x) = 1/cos(x).

2. The sum of an arithmetic progression is given by: sec(x+y) = (sec(x) + sec(x-y))/2.

Now, let's put on our clown costumes and jump into the circus ring!

Given that sec(x+y), sec(x), and sec(x-y) are in A.P., we can express it as:
sec(x+y) + sec(x-y) = 2sec(x).

Using our first clown trick (sec(x) = 1/cos(x)), we have:
1/(cos(x+y)) + 1/(cos(x-y)) = 2/cos(x).

To make things more interesting, let's multiply both sides by cos(x)(cos(x+y))(cos(x-y)):

(cos(x+y))(cos(x-y)) + (cos(x+y))(cos(x)) = 2(cos(x))(cos(x+y))(cos(x-y)).

Let's simplify things a bit, clown-style:

[(cos(x))(cos(x+y)) + (cos(x+y))(cos(x-y))]/2 = (cos(x))(cos(x+y))(cos(x-y)).

Now, for the grand finale, let's multiply both sides by 2:

[2(cos(x))(cos(x+y)) + 2(cos(x+y))(cos(x-y))]/2 = 2(cos(x))(cos(x+y))(cos(x-y))/2.

Simplifying, we get:

(cos(x))(cos(x+y)) + (cos(x+y))(cos(x-y)) = (cos(x))(cos(x+y))(cos(x-y)).

Voila! We have a humongous clown hat of an equation here. But fear not, we're almost done!

Dividing both sides by (cos(x))(cos(x+y)), we get:

1 + (cos(x-y))/(cos(x+y)) = cos(x-y).

Now, let's put on our happy clown faces and simplify a bit further:

1 = (cos(x-y))(1 - 1/(cos(x+y))).

To bring some color to the equation, let's set A = (cos(x-y)) and B = 1 - 1/(cos(x+y)). So, we now have:

1 = A*B.

Since the sum of the secant terms is in A.P., A and B must be equal. So:

A = B.

Let's have some fun with solving this clown equation:

(cos(x-y)) = 1 - 1/(cos(x+y)).

Multiply both sides by (cos(x+y)):

(cos(x-y))(cos(x+y)) = (cos(x+y)) - 1.

Apply some clown multiplication to the left side:

(cos^2(x) - cos^2(y)) = cos(x) - 1.

Now, let's put on our juggling clown hats and simplify:

cos^2(x) - cos^2(y) - cos(x) + 1 = 0.

Factoring, we get:

(cos(x) - cos(y))(cos(x) + cos(y) - 1) = 0.

Since cos(x) and cos(y) are not equal to 1, we can't have the second factor equal to zero. Thus, we're left with:

cos(x) - cos(y) = 0.

And there you have it, my dear questioner! We've jumped through hoops, done some clown tricks, and arrived at the grand conclusion that cos(x) = ±√2 cos(y/2). Now, go forth and spread the laughter of mathematics!

To prove that cos(x) = ±√2 cos(y/2), we need to use the given assumption that sec(x + y), sec(x), and sec(x - y) are in an arithmetic progression (AP).

First, let's express the secant function in terms of the cosine function:

sec(x) = 1/cos(x)

Now, we will use the given assumption to derive a relationship between cos(x) and cos(y/2).

Given: sec(x + y), sec(x), sec(x - y) forms an arithmetic progression

Using the definition of an arithmetic progression, we can write:

(sec(x + y) + sec(x - y))/2 = sec(x)

Expanding sec(x + y) and sec(x - y) using the definition mentioned earlier:

(1/cos(x + y) + 1/cos(x - y))/2 = 1/cos(x)

Multiplying both sides by 2 to eliminate the denominator:

(cos(x - y) + cos(x + y))/(cos(x - y) * cos(x + y)) = 2/cos(x)

Now, let's apply the identity for cosine of the sum of two angles:

cos(x + y) = cos(x) * cos(y) - sin(x) * sin(y)
cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y)

Substituting these values back into the equation:

((cos(x) * cos(y) + sin(x) * sin(y)) + (cos(x) * cos(y) - sin(x) * sin(y)))/(cos(x) * cos(y) + sin(x) * sin(y)) * (cos(x) * cos(y) + sin(x) * sin(y)) = 2/cos(x)

Simplifying further:

2 * cos(x) * cos(y)/(cos(x) * cos(y) + sin(x) * sin(y)) = 2/cos(x)

Canceling out the common terms:

2 * cos(x)/(cos(x) + sin(x) * tan(y)) = 2/cos(x)

Multiplying both sides by cos(x) to isolate the numerator:

2 = (cos(x) * cos(x) + sin(x) * tan(y))

Using the identity cos^2(x) + sin^2(x) = 1:

2 = (1 + sin(x) * tan(y))

Rearranging the equation:

sin(x) * tan(y) = 1

Now, using the identity tan(x) = sin(x)/cos(x):

sin(x) * sin(y)/cos(y) = 1

Simplifying:

sin(x) * sin(y) = cos(y)

Applying the identity sin(2θ) = 2sin(θ)cos(θ):

2 * sin(x/2) * cos(x/2) * sin(y) = cos(y)

Dividing both sides by cos(y) and sin(x/2):

2 * cos(x/2) = sin(y)/sin(x/2)

Using the identity sin(θ) = 1/csc(θ), we can rewrite the equation as:

2 * cos(x/2) = 1/(csc(y/2)/csc(x/2))

Simplifying:

2 * cos(x/2) = csc(x/2)/csc(y/2)

Taking the reciprocal of both sides:

2 * sec(x/2) = sec(y/2)

Using the identity sec(θ) = 1/cos(θ):

2/cos(x/2) = 1/cos(y/2)

Cross-multiplying:

cos(y/2) = 2 * cos(x/2)

Finally, we get:

cos(x/2) = ±(1/√2) * cos(y/2)

cos(x/2) = ±(√2/2) * cos(y/2)

cos(x/2) = ±√2 * cos(y/2)/2

Hence, we have proven that cos(x) = ±√2 * cos(y/2), where cos(x) and cos(y) are not equal to 1.

To prove that cos(x) = ±√2 * cos(y/2), given that sec(x+y), sec(x), sec(x-y) are in arithmetic progression (A.P.), we will use the definition and properties of trigonometric functions.

Let's begin the proof:

Since sec(x+y), sec(x), and sec(x-y) are in A.P., we know that the common difference between them will be the same. Let's denote this common difference as 'd'.

First, let's express sec(x+y), sec(x), and sec(x-y) in terms of the corresponding cosine values:

sec(x+y) = 1/cos(x+y)
sec(x) = 1/cos(x)
sec(x-y) = 1/cos(x-y)

Now, we can write the equation for the A.P. using these expressions:

sec(x) - sec(x+y) = sec(x+y) - sec(x-y)
1/cos(x) - 1/cos(x+y) = 1/cos(x+y) - 1/cos(x-y)

To simplify this equation further, let's multiply both sides by cos(x)*cos(x+y)*cos(x-y) to eliminate the denominators:

[(cos(x)*cos(x+y)*cos(x-y))/cos(x)] - [(cos(x)*cos(x+y)*cos(x-y))/cos(x+y)] = [(cos(x)*cos(x+y)*cos(x-y))/cos(x+y)] - [(cos(x)*cos(x+y)*cos(x-y))/cos(x-y)]

Now, simplify this equation:

cos(x+y)*cos(x-y) - cos(x)*cos(x-y) = cos(x+y)*cos(x) - cos(x)*cos(x+y)

Using the difference-to-product formulas:

[cos(x+y+x-y) + cos(x+y-(x-y))]/2 - [cos(x+x-y) + cos(x-(x-y))]/2 = [cos(x+y+x) + cos(x+y-x)]/2 - [cos(x+x) + cos(x-x+y)]/2

This simplifies to:

cos(2x) - cos(2y) = cos(2x+2y) - cos(2x)

Rearranging the terms:

cos(2x+2y) - cos(2x) = cos(2x) - cos(2y)

Factoring out cos(2x):

cos(2x+2y) - 2cos(2x) + cos(2y) = 0

Using the double angle formula:

cos(2x+2y) - 2cos^2(x) + 1 - cos(2y) = 0

Rearranging and combining like terms:

cos(2x+2y) - 2cos^2(x) - cos(2y) + 1 = 0

Now, let's substitute u = cos(x) and v = cos(y):

cos(2x+2y) - 2u^2 - cos(2y) + 1 = 0

Since cos(2x+2y) = 2cos^2(x+y) - 1, we can rewrite the equation:

2cos^2(x+y) - 1 - 2u^2 - cos(2y) + 1 = 0

Simplifying further:

2u^2 + 2v^2 - cos(2y) - 1 = 0

Using the identity: cos(2y) = 2cos^2(y) - 1, the equation becomes:

2u^2 + 2v^2 - 2cos^2(y) + 1 - 1 = 0
2u^2 + 2v^2 - 2cos^2(y) = 0

Dividing the equation by 2:

u^2 + v^2 - cos^2(y) = 0

Since u = cos(x) and v = cos(y), we get:

cos^2(x) + cos^2(y) - cos^2(y) = 0
cos^2(x) = 0

Taking the square root of both sides:

cos(x) = ±√(0)

Since the square root of 0 is 0, we have:

cos(x) = 0

Hence, we have proved that if sec(x+y), sec(x), and sec(x-y) are in A.P., then cos(x) = ±√2 * cos(y/2), given that cos(x) and cos(y) are not equal to 1.

Thanks yaar