Question

Prove that: cos 9°+sin 9° /cos 9° - sin 9° = tan 37°

Answer 1

Correct Question:

Prove that:

$\rm \hookrightarrow \dfrac{ \cos(9{}^{\circ}) + \sin(9{}^{\circ}) }{ \cos(9{}^{\circ}) - \sin(9{}^{\circ}) } = \tan(54{}^{\circ})$

Solution:

Taking Left Hand Side, we get:

$\rm = \dfrac{ \cos(9{}^{\circ}) + \sin(9{}^{\circ}) }{ \cos(9{}^{\circ}) - \sin(9{}^{\circ}) }$

Dividing both numerator and denominator by cos(9°), we get:

$\rm = \dfrac{ \frac{1}{ \cos(9{}^{\circ}) } \{\cos(9{}^{\circ}) + \sin(9{}^{\circ}) \} }{ \frac{1}{ \cos(9{}^{\circ}) } \{ \cos(9{}^{\circ}) - \sin(9{}^{\circ}) \} }$

$\rm = \dfrac{1 + \tan(9{}^{\circ}) }{1 - \tan(9{}^{\circ}) }$

Can be written as:

$\rm = \dfrac{1 + \tan(9{}^{\circ}) }{1 - 1 \times \tan(9{}^{\circ}) }$

1 can be written as tan(45°). So, we get:

$\rm = \dfrac{ \tan(45{}^{\circ}) + \tan(9{}^{\circ}) }{1 - \tan(45{}^{\circ}) \times \tan(9{}^{\circ}) }$

We know that:

$\rm \longrightarrow \tan( \alpha + \beta ) = \dfrac{ \tan( \alpha ) + \tan( \beta) }{1 - \tan( \alpha ) \tan( \beta ) }$

Using this result, we get:

$\rm = \tan(45{}^{\circ} + 9{}^{\circ})$

$\rm = \tan(54{}^{\circ})$

$\rm = RHS$

Therefore:

$\rm \hookrightarrow \dfrac{ \cos(9{}^{\circ}) + \sin(9{}^{\circ}) }{ \cos(9{}^{\circ}) - \sin(9{}^{\circ}) } = \tan(54{}^{\circ})$

Hence Proved..!!

Additional Information:

$\rm1. \: \sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$

$\rm2.\: \cos(\alpha+\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$

$\rm 3.\: \tan(\alpha+\beta) = \dfrac{ \tan( \alpha) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }$

$\rm 4. \: \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$

$\rm 5.\: \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$

$\rm 6.\: \tan(\alpha - \beta) = \dfrac{ \tan( \alpha) - \tan( \beta ) }{1 + \tan( \alpha ) \tan( \beta ) }$