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Basic Differentiation Rule

If xy = cos...

Question

If $\frac{x}{y} = \frac{cos A}{cos B}$ where $A \neq B$ , then

A

tan \frac{A + B}{2} = \frac{x tan A + y tan B}{x + y}

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B

tan \frac{A - B}{2} = \frac{x tan A - y tan B}{x + y}

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C

\frac{sin (A + B)}{sin (A - B)} = \frac{y sin A + x sin B}{y sin A - x sin B}

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D

x cos A + y cos B = 0

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Solution

The correct options are
A $tan \frac{A + B}{2} = \frac{x tan A + y tan B}{x + y}$
B $\frac{sin (A + B)}{sin (A - B)} = \frac{y sin A + x sin B}{y sin A - x sin B}$
D $tan \frac{A - B}{2} = \frac{x tan A - y tan B}{x + y}$
Let $\frac{cos A}{x} = \frac{cos B}{y} = k$
$\Rightarrow \frac{cos A}{sin A} = \frac{x k}{sin A}$ and $\frac{cos B}{sin B} = \frac{y k}{sin B}$
Therefore,
$\frac{x tan A + y tan B}{x + y} = \frac{sin A + sin B}{cos A + cos B} = tan (\frac{A + B}{2})$
$\frac{x tan A - y tan B}{x + y} = \frac{sin A - sin B}{cos A + cos B} = tan (\frac{A - B}{2})$
$\frac{y sin A + x sin B}{y sin A - x sin B} = \frac{sin A cos B + cos A sin B}{sin A cos B - cos A sin B} = \frac{sin (A + B)}{sin (A - B)}$

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Similar questions

\frac{x}{y} = \frac{cos A}{cos B}

\frac{x tan A + y tan B}{x + y} =

tan A = \frac{x sin B}{1 - x cos B}

tan B = \frac{y sin A}{1 - y cos A}

, then the value of

\frac{sin A}{sin B}

for all permissible values of

A, B

tan A = \frac{x sin B}{1 - x cos B}

tan B = \frac{y sin A}{1 - y cos A}

\frac{sin A}{sin B} =

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

\frac{sin (A + B)}{sin (A - B)} = \frac{x + y}{x - y}

\frac{tan A}{tan B}

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