Trigonometric Functions

(√3 - 1)/(2√2)

Step 1: Express 15° = 45° - 30° and use the identity sin(A - B) = sin A cos B - cos A sin B. Step 2: sin 15° = sin 45° cos 30° - cos 45° sin 30°. Step 3: = (1/√2)(√3/2) - (1/√2)(1/2). Step 4: = √3/(2√2) - 1/(2√2) = (√3 - 1)/(2√2). Note: This can also be written as (√6 - √2)/4, but (√3-1)/(2√2) is the direct simplified form. Option C is incorrect as written.

2 - √3

Step 1: Write 13π/12 = π + π/12, so tan(13π/12) = tan(π/12) (since tan(π + x) = tan x). Step 2: Write π/12 = π/4 - π/6 and apply tan(A - B) = (tan A - tan B)/(1 + tan A tan B). Step 3: tan(π/12) = (tan π/4 - tan π/6)/(1 + tan π/4 · tan π/6) = (1 - 1/√3)/(1 + 1/√3). Step 4: Multiply numerator and denominator by √3: = (√3 - 1)/(√3 + 1). Step 5: Rationalise: (√3-1)²/((√3)²-1²) = (4-2√3)/2 = 2 - √3.

a/b = tan x / tan y

Step 1: Expand sin(x+y) = sin x cos y + cos x sin y and sin(x-y) = sin x cos y - cos x sin y. Step 2: So (sin x cos y + cos x sin y)/(sin x cos y - cos x sin y) = (a+b)/(a-b). Step 3: Divide numerator and denominator of LHS by cos x cos y: (tan x + tan y)/(tan x - tan y) = (a+b)/(a-b). Step 4: By componendo-dividendo or direct comparison: a corresponds to tan x and b to tan y. Step 5: Therefore a/b = tan x / tan y.

cos 2x

Step 1: We know cos(A + B) = cos A cos B - sin A sin B. Step 2: Put A = B = x: cos 2x = cos²x - sin²x. Step 3: This is the fundamental double angle formula for cosine. Step 4: cos 2x = cos²x - sin²x is the direct identity. The options '1 - 2sin²x' and '2cos²x - 1' are also equal to cos 2x, but they are not equal to cos²x - sin²x as individual standalone expressions — they are all equivalent forms. The complete identity is cos 2x = cos²x - sin²x = 1 - 2sin²x = 2cos²x - 1.