If x y = ab , then a x + b y equals

The ratio x y = ab means a and b are proportional to x and y : write a = k x and b = k y for some constant k . Then a x + b y = k x x x + k y y y = k( x + y), and a + b = k( x + y). Using the sum-to-product identities x + y = 2 x+y2 x-y2 and x + y = 2 x+y2 x-y2 , a x + b ya + b = x + y x + y = 2 x+y2 x-y22 x+y2 x-y2 = x+y2. Therefore a x + b y = (a + b) x+y2 . Answer: B.

Trigonometric Identities Difficulty 4

TMUA practice: Trig Identities, problem 4

A TMUA-calibre trig identities problem at difficulty 4 of 5, with the full worked solution.

The question

\dfrac{\cos x}{\cos y} = \dfrac{a}{b}

, then

a\tan x + b\tan y

equals

(a + b)\cot\dfrac{x + y}{2}

(a + b)\tan\dfrac{x + y}{2}

(a + b)\left(\tan\dfrac{x}{2} + \tan\dfrac{y}{2}\right)

(a + b)\left(\cot\dfrac{x}{2} + \cot\dfrac{y}{2}\right)

The correct answer is highlighted. B

Worked solution

The ratio $\dfrac{\cos x}{\cos y} = \dfrac{a}{b}$ means $a$ and $b$ are proportional to $\cos x$ and $\cos y$ : write $a = k\cos x$ and $b = k\cos y$ for some constant $k$ .

Then

a\tan x + b\tan y = k\cos x\cdot\frac{\sin x}{\cos x} + k\cos y\cdot\frac{\sin y}{\cos y} = k(\sin x + \sin y),

and

$a + b = k(\cos x + \cos y).$

Using the sum-to-product identities $\sin x + \sin y = 2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2}$ and $\cos x + \cos y = 2\cos\tfrac{x+y}{2}\cos\tfrac{x-y}{2}$ ,

\frac{a\tan x + b\tan y}{a + b} = \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2}}{2\cos\tfrac{x+y}{2}\cos\tfrac{x-y}{2}} = \tan\frac{x+y}{2}.

Therefore $a\tan x + b\tan y = (a + b)\tan\dfrac{x+y}{2}$ .

Answer: B.