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If \sin A + \sin B = p and \cos A - \cos B = q then \tan \frac{A-B}{2} is equal to

 

  • Option 1)

    \frac{-p}{q}

  • Option 2)

    \frac{-q}{p}

  • Option 3)

    \sqrt{p^{2}+q^{2}}

  • Option 4)

    None of these

 

Answers (1)

 

Transformation Formulae -

transformation formulae 1

- wherein

These formula are also called C-D formulae.

 

\sin A + \sin B = p

\cos A-\cos B = q
\Rightarrow 2\sin \frac{A+B}{2}\cdot \cos \frac{A-B}{2} = p

-2\sin \frac{A+B}{2} \sin \frac{A-B}{2} = q

\therefore \tan \frac{A-B}{2} = \frac{-q}{p}


Option 1)

\frac{-p}{q}

This solution is incorrect.

Option 2)

\frac{-q}{p}

This solution is correct.

Option 3)

\sqrt{p^{2}+q^{2}}

This solution is incorrect.

Option 4)

None of these

This solution is incorrect.

Posted by

Vakul

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