s i n A − s i n B c o s A + c o s B + c o s A − c o s B s i n A + s i n B = Options: A . &nbsp - 1 B . &nbsp 0 C . &nbsp 1 D . &nbsp 2

Answer: Option B : B s i n A − s i n B c o s A + c o s B + c o s A − c o s B s i n A + s i n B = ( s i n A − s i n B ) ( s i n A + s i n B ) + ( c o s A − c o s B ) ( c o s A + c o s B ) ( c o s A + c o s B ) ( s i n A + s i n B ) = s i n 2 A − s i n 2 B + c o s 2 A − c o s 2 B ( c o s A + c o s B ) ( s i n A + s i n B ) = s i n 2 A + c o s 2 A − ( s i n 2 B ‘ + c o s 2 B ) ( c o s A + c o s B ) ( s i n A + s i n B ) = 1 − 1 ( c o s A + c o s B ) ( s i n A + s i n B ) = 0

sinA−sinBcosA+cosB+cosA−cosBsinA+sinB=

Question

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$

Options:

A . - 1

B . 0

C . 1

D . 2

Answer: Option B
:
B

$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

= $\frac{(s i n A - s i n B) (s i n A + s i n B) + (c o s A - c o s B) (c o s A + c o s B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A - s i n^{2} B + c o s^{2} A - c o s^{2} B}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A + c o s^{2} A - (s i n^{2} B ‘ + c o s^{2} B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{1 - 1}{(c o s A + c o s B) (s i n A + s i n B)}$
= 0