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

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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

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