Maths

Given the equation 15 sin? + 10 cos? = 6.
Divide by cos? : 15 tan? + 10 = 6 sec?
Using sec²? = 1 + tan²? , we get sec? = (1 + tan²? )² = 1 + 2tan²? + tan?
15 tan? + 10 = 6 (1 + 2tan²? + tan? ).
15 tan? + 10 = 6 + 12tan²? + 6tan?
9 tan? - 12 tan²? + 4 = 0.
This is a quadratic in tan²? : (3 tan²? - 2)² = 0.
? 3 tan²? = 2? tan²? = 2/3.
From this, we find sin²? and cos²? If tan²? = 2/3, then sin²? = 2/5 and cos²? = 3/5.
Also, sec²? = 1 + tan²? = 5/3 and cosec²? = 1 + cot²? = 1 + 3/2 = 5/2.
The expression to evaluate is 27 sec? + 8 cosec? = 27 (sec²? )³ + 8 (cosec²? )³.
= 27 (5/3)³ + 8 (5/2)³ = 27 (125/27) + 8 (125/8) = 125 + 125 = 250.

Given vectors a? and b? such that |a? | = |b? | and a? ⋅ b? = 0 (they are orthogonal).
The problem implies |a? |=|b? |=1.
Let c? = a? + b? + a? x b?
To find the magnitude of c? , we calculate |c? |²:
|c? |² = c? ⋅ c? = (a? + b? + a? x b? ) ⋅ (a? + b? + a? x b? ).
This expands to |a? |² + |b? |² + |a? x b? |² because all other dot products are zero (e.g., a? ⋅ b? = 0, a? ⋅ (a? x b? ) = 0).
|a? x b? |² = (|a? |b? |sin (90°)² = |a? |²|b? |².
So, |c? |² = |a? |² + |b? |² + |a? |²|b? |² = 1² + 1² + 1²*1² = 3.
∴ |c? | = √3.
To find the angle θ between c? and a? , we compute their dot product:
c? ⋅ a? = (a? + b? + a? x b? ) ⋅ a? = a? ⋅ a? + b? ⋅ a? + (a? x b? ) ⋅ a?
c? ⋅ a? = |a? |² + 0 + 0 = 1.
Now, cos θ = (c? ⋅ a? ) / (|c? |a? |) = 1 / (√3 * 1) = 1/√3.
⇒ θ = cos? ¹ (1/√3).

Given functions f (x) = (x-2)/ (x-3) and g (x) = 2x-3.
First, find the inverse functions f? ¹ (x) and g? ¹ (x).
For f? ¹ (x): y = (x-2)/ (x-3) ⇒ y (x-3) = x-2 ⇒ xy - 3y = x-2 ⇒ xy-x = 3y-2 ⇒ x (y-1) = 3y-2 ⇒ x = (3y-2)/ (y-1). So, f? ¹ (y) = (3y-2)/ (y-1).
For g? ¹ (x): y = 2x-3 ⇒ y+3 = 2x ⇒ x = (y+3)/2. So, g? ¹ (y) = (y+3)/2.
We are given f? ¹ (x) + g? ¹ (x) = 13/2.
(3x-2)/ (x-1) + (x+3)/2 = 13/2.
Multiply by 2 (x-1): 2 (3x-2) + (x+3) (x-1) = 13 (x-1).
6x - 4 + x² + 2x - 3 = 13x - 13.
x² + 8x - 7 = 13x - 13.
x² - 5x + 6 = 0.
(x-2) (x-3) = 0.
The possible values of x are 2 and 3. Note that x=3 is not in the domain of the original f (x), and x=1 is not in the domain of f? ¹ (x). However, the equation is solved for x.
The sum of possible values is 2 + 3 = 5.

The equation is for a hyperbola: x²/4 - y²/2 = 1.
The eccentricity e is given by e = √ (1 + b²/a²) = √ (1 + 2/4) = √6/2.
The focus F is at (ae, 0), which is (2 * √6/2, 0) = (√6, 0).
The equation of the tangent at a point P (x? , y? ) is xx? /a² - yy? /b² = 1.
The equation of the tangent at P is given as x - (√6 y)/2 = 1.
This tangent cuts the x-axis (y=0) at x=1, so Q is (1, 0).
The latus rectum is the line x = ae = √6.
To find the point R where the tangent intersects the latus rectum, we substitute x=√6 into the tangent equation:
√6 - (√6 y)/2 = 1 ⇒ √6 - 1 = (√6 y)/2 ⇒ y = 2 (√6 - 1)/√6.
The vertices of the triangle QFR are Q (1,0), F (√6,0), and R (√6, 2 (√6 - 1)/√6).
The base QF has length √6 - 1.
The height is the y-coordinate of R.
Area of ΔQFR = 1/2 * base * height = 1/2 * (√6 - 1) * [2 (√6 - 1)/√6].
Area = (√6 - 1)² / √6 = (6 - 2√6 + 1) / √6 = (7 - 2√6) / √6.

A relation R is defined as ARB if PAP? ¹ = B for a non-singular matrix P.

·       Reflexive: ARA requires PAP? ¹ = A. This holds if P is the identity matrix I, as IAI? ¹ = A. Assuming P can be I, the relation is reflexive.

·       Symmetric: We need to show that if ARB, then BRA.
ARB ⇒ PAP? ¹ = B.
To get the reverse, we need to express A in terms of B.
From PAP? ¹ = B, pre-multiply by P? ¹ and post-multiply by P:
P? ¹ (PAP? ¹)P = P? ¹BP ⇒ A = P? ¹BP. This shows BRA where the matrix is P? ¹. Thus, the relation is symmetric.

·       Transitive: We need to show that if ARB and BRC, then ARC.
ARB ⇒ PAP? ¹ = B — (I)
BRC ⇒ PBP? ¹ = C — (II)
Substitute B from (I) into (II):
P (PAP? ¹)P? ¹ = C
⇒ (P²)A (P? ¹)² = C
⇒ (P²)A (P²)? ¹ = C
This is in the form QAQ? ¹ = C with Q = P². This implies ARC. Thus, the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.

Let x = m(a + λb).
Given m(a + λb) ⋅ (3i + 2j - k) = 0, which leads to λ = -3/8.

The projection of vector x on vector a is given by x ⋅ â, where â is the unit vector of a.
Projection = (x ⋅ a) / |a| = 17√6 / 2
x ⋅ a = (m(a + λb)) ⋅ a = m(a ⋅ a + λ(b ⋅ a)) = m(|a|^2 + λ(b ⋅ a))

The provided text simplifies this to:
m(6 - 3/8 * (-1)) = 17√6 / 2
m * (51/8) = 17 * 6 / 2 (The text seems to have a typo 17x6/2 instead of 17√6 / 2)
Assuming it is 17 * 6 / 2, m * 51/8 = 51, so m = 8.

x = 8(a + (-3/8)b) = 8a - 3b
x = 8( (13/8)i - (14/8)j + (11/8)k ) (The vectors a and b are not fully defined in the provided text)
The final vector is given as x = 13i - 14j + 11k.

|x|^2 = 13^2 + (-14)^2 + 11^2 = 169 + 196 + 121 = 486.

The problem provides an equation involving the coordinates (α, β, γ) of a point P:
((α + β + γ) / √3)^2 + ((α - nγ) / √(l^2 + n^2))^2 + ((α - 2β + γ) / √6)^2 = 9

The locus of P(α, β, γ) is given by replacing (α, β, γ) with (x, y, z):
((x + y + z) / √3)^2 + ((lx - nz) / √(l^2 + n^2))^2 + ((x - 2y + z) / √6)^2 = 9

This represents the equation of an ellipsoid. The text proceeds by comparing coefficients. By expanding the equation, the coefficients of x^2, y^2, z^2, and cross-product terms are collected. From the given conditions:
Coefficient of x^2: 1/3 + l^2 / (l^2 + n^2) + 1/6 = 1
Coefficient of y^2: 1/3 + 0 + 4/6 = 1
Coefficient of z^2: 1/3 + n^2 / (l^2 + n^2) + 1/6 = 1
Coefficient of xz: -2ln / (l^2 + n^2) + 2/6 = 0
(The document seems to simplify this differently).

The text provides a shortcut:
1/3 + (l^2 / (l^2 + n^2)) + 1/2 = 1 (Combining 1/3 and 1/6 to 1/2)
And also, (ln / (l^2 + n^2)) = 0. This implies ln = 0.

If ln = 0, then from l^2/(l^2+n^2) = 1/2, this means 2l^2 = l^2+n^2, so l^2 = n^2, which implies l = ±n.
The text states l^2 + n^2 - 2ln = 0, which simplifies to (l - n)^2 = 0, therefore l - n = 0 or l = n.