(cos9 + \sin9 )/(cos9 - sin9) = tan54 Prove the above question​

Question

`(cos9 + \sin9 )/(cos9 - sin9) = tan54`
Prove the above question​

Answer 1

Solution:

To Prove :

`\sf (\cos9\degree + \sin9\degree )/(\cos9\degree- \sin9\degree) = \tan54\degree \\ \\`

`\sf LHS = (\cos9\degree + \sin9\degree )/(\cos9\degree- \sin9\degree) \\ \\`

`\sf RHS = \tan54\degree \\ \\`

Solving for LHS :

Divide numerator and denominator by cos9°

`\longrightarrow \: \: \sf\frac { ( \cos9\degree)/( \cos9\degree) + { (\sin9)/( \cos9) }}{ ( \cos9\degree )/( \cos9\degree) - { ( \sin9\degree)/( \cos9\degree)}} \\ \\`

`\longrightarrow \: \: \sf\frac{1 + { \frac {\sin9\degree}{ \cos9\degree}} }{1 - { ( \sin9\degree)/( \cos9\degree) }} \\ \\`

`\longrightarrow \: \: \sf (1 + \tan9\degree)/(1 - \tan9\degree) \\ \\`

`\longrightarrow \: \: \sf{ ( \tan45 \degree + \tan9\degree)/(1 - \tan45 \degree \tan9 \degree)} \: \: \: \: \: \: \: \sf \red{\bigg( \tan(A + B) = ( \tan A + \tan B)/(1 - \tan A \tan B) \bigg)} \\ \\`

`\longrightarrow \: \: \sf \tan(45\degree + 9\degree) \\ \\`

`\longrightarrow \: \: \sf \tan54 \degree`

LHS = RHS

Hence, proved!!