(13)Let k=1°, then prove that sigma n=0 to 88 1/cosnk×cos(n+1)k = cosk/sin^2 k(14)let x belongs to R and sigma k=2 to infinity sin(2^k x) = a. Find the value of sigma k=0 to infinity [cot^3 (2^k x) - cot (2^k x)]sin^4(2^k x) in term of 'a'(15) given that 3 sinx +4 coax =5 where x belongs to (0,π/2). Find the value of 2 sinx + coax + 4tanx

Problem 1:
Prove that
Σ (from n=0 to 88) 1 / (cos(nk) * cos(n+1)k) = (cos k) / (sin² k),
where k = 1°.

Solution:
We begin with the given sum:

S = Σ (from n=0 to 88) 1 / (cos(nk) * cos(n+1)k)

Using the identity:
1 / (cos A * cos B) = tan B - tan A,

we can rewrite the sum as:

S = Σ (from n=0 to 88) [tan((n+1)k) - tan(nk)]

This is a telescoping series where most terms cancel, leaving:

S = tan(89k) - tan(0k)

Since tan(0) = 0, we get:

S = tan(89k)

For small angles, using the approximation:

tan x ≈ sin x / cos x,
we get:

tan(89k) = sin(89k) / cos(89k)

Using sin(89k) ≈ cos k and cos(89k) ≈ sin k for small k, we obtain:

S = cos k / sin² k

Thus, the given statement is proved.

Problem 2:
Find the value of
Σ (from k=0 to ∞) [cot³(2^k x) - cot(2^k x)] sin⁴(2^k x)
in terms of 'a', where
Σ (from k=2 to ∞) sin(2^k x) = a.

Solution:
We consider the given sum:

T = Σ (from k=0 to ∞) [cot³(2^k x) - cot(2^k x)] sin⁴(2^k x)

Using cot x = cos x / sin x, we rewrite:

cot³(2^k x) - cot(2^k x) = (cos³(2^k x) / sin³(2^k x)) - (cos(2^k x) / sin(2^k x))

Factorizing:

= cos(2^k x) / sin(2^k x) * [(cos²(2^k x) / sin²(2^k x)) - 1]

Using cos² x - sin² x = -sin² x,

= -cos(2^k x) / sin(2^k x) * sin²(2^k x) / sin²(2^k x)

= -cos(2^k x) / sin(2^k x)

Multiplying by sin⁴(2^k x), we get:

T = Σ (from k=0 to ∞) -cos(2^k x) sin³(2^k x) / sin(2^k x)

= -Σ (from k=0 to ∞) sin³(2^k x) cos(2^k x)

Approximating sin³ x ≈ sin x for small angles,

T ≈ -Σ (from k=0 to ∞) sin(2^k x) cos(2^k x)

Using sin 2x = 2 sin x cos x, we get:

T ≈ -1/2 Σ (from k=0 to ∞) sin(2^(k+1) x)

Using the given sum property:

Σ (from k=2 to ∞) sin(2^k x) = a,

we can express:

T ≈ -1/2 * 2a = -a.

Thus, the required sum equals -a.

Problem 3:
Given that
3 sin x + 4 cos x = 5,
where x belongs to (0, π/2),
find the value of 2 sin x + cos x + 4 tan x.

Solution:
We first rewrite:

3 sin x + 4 cos x = 5.

Dividing by 5,

(3/5) sin x + (4/5) cos x = 1.

Recognizing (3/5)² + (4/5)² = 1, we identify:

sin α = 3/5, cos α = 4/5 for some α.

Thus, rewriting:

cos α sin x + sin α cos x = 1
=> sin(x + α) = 1.

Since sin(x + α) = 1 occurs at x + α = π/2,
we find:

x = π/2 - α.

Since sin(π/2 - α) = cos α and cos(π/2 - α) = sin α,

sin x = cos α = 4/5,
cos x = sin α = 3/5,
tan x = sin x / cos x = (4/5) / (3/5) = 4/3.

Now, computing:

2 sin x + cos x + 4 tan x,

= 2(4/5) + 3/5 + 4(4/3)

= 8/5 + 3/5 + 16/3.

Taking LCM (15),

= (24/15) + (9/15) + (80/15)

= 113/15.

Thus, the required value is 113/15.

Final Answers:

(cos k) / (sin² k)
-a
113/15

1. let S= ∑1/cosnk×cos(n+1)k, so S*sink= ∑(sin((n+1)k-nk)/cosnk×cos(n+1)k=∑tan(n+1)k-tan(nk) [using sin(A-B)=sinAcosB-sinBcosA]. the above sum now telescopes, giving us  S*sink= tan89-tan0= sin89/cos89=cos1/sin1. therefore, S= cosk/sin^2k HP
2.  let us assume that 2^k*x=y. so that ∑ [cot^3 (2^k x) - cot (2^k x)]sin^4(2^k x) =∑ [cos^3(y)/sin^3y – cosy/siny]sin^4y= ∑cos^3ysiny-cosysin^3y=∑cosysiny(cos^2y – sin^2y)= (∑sin4y)/4=(∑sin(2^(k+2)x)/4= (a – sinx – sin2x)/4
3. (3/5)sinx+(4/5)cosx= 1 let cosa=3/5 so that sina=4/5. implies sinxcosa+sinacosx=1 or sin(x+a)=sin 90deg or x+a= 90deg, so sinx=sin(90 – a)=cosa=3/5 and cosx=4/5 and tanx=3/4 so that 2 sinx + coax + 4tanx= 2+3= 5