Question

Question :-

$\sf\dfrac{cos2A - cos2B}{sin2A + sin2B}$ =

a) tan(A + B)
b) tan(B - A)
c) tan(A + B)/2
d) - tan(A + B)/2​

Answer 1

Answer:

b. tan(B-A)

Step-by-step explanation:

Used x, y instead of A, B

$\frac{ \cos(2x) - \cos(2y) }{ \sin(2x) + \sin(2y) } \\ = \frac{ - 2 \sin( \frac{2x + 2y}{2} ) \sin( \frac{2x - 2y}{2} ) }{2 \sin( \frac{2x + 2y}{2} ) \cos( \frac{2x - 2y}{2} ) } \\ \frac{ - \sin( x - y) }{ \cos( x - y) } = - \tan(x - y)$

-tan(x-y) =tan(y-x)

Answer 2

EXPLANATION.

$\sf \implies \dfrac{cos2A - cos2B}{sin2A + sin2B}$

As we know that,

Formula of :

⇒ cos C - cos D = 2 sin(C + D)/2. sin(D - C)/2.

⇒ sin C + sin D = 2 sin(C + D)/2. cos(C - D)/2.

Using this formula in equation, we get.

$\sf \implies \dfrac{2 \ sin \bigg(\dfrac{2A + 2B}{2}\bigg) . sin \bigg(\dfrac{2B - 2A}{2} \bigg)}{2 \ sin \bigg(\dfrac{2A + 2B}{2} \bigg). cos \bigg(\dfrac{2A - 2B}{2}\bigg) }$

$\sf \implies \dfrac{2 \ sin (A + B) . sin(B - A)}{2 \ sin(A + B) . cos(A - B)}$

$\sf \implies \dfrac{sin(B - A)}{cos(A - B)}$

As we know that,

⇒ cos(-x) = cos(x)

Using this formula in equation, we get.

$\sf \implies \dfrac{sin(B - A)}{cos(B - A)} = tan (B - A)$

Hence, option [B] is correct answer.

                                                                                                                       

MORE INFORMATION.

Sum & Difference formula.

(1) = sin(A ± B) = sin(A).cos(B) ± cos(A).sin(B).

(2) = cos(A + B) = cos(A).cos(B) - sin(A).sin(B).

(3) = cos(A - B) = cos(A).cos(B) + sin(A).sin(B).

(4) = tan(A + B) = tan(A) + tan(B)/1 - tan(A).tan(B).

(5) = tan(A - B) = tan(A) - tan(B)/ 1 + tan(A).tan(B).

(6) = cot(A + B) = cot(A).cot(B) - 1/cot(B) + cot(A).

(7) = cot(A - B) = cot(A).cot(B) + 1/cot(B) - cot(A).