Question

Prove that tan 9 = (cos 36 - sin 36) / (cos 36 + sin 36) Using L.H.S. Answer the Brainiest for this question.

Answer 1

Answer:

L.H.S. = R.H.S.

Step-by-step explanation:

L.H.S. = (cos 36 - sin 36) / (cos 36 + sin 36)  

        = (cos 36 - cos 54) / (cos 36 +cos 54)

        = 2 sin 45 sin 9 / 2 cos 45 cos 9

        =  sin 9 /  cos 9

        =  tan 9

         = R.H.S.

Answer 2

AnswEr :

• To Prove :

$\tan(9) = \dfrac{( \cos(36) - \sin(36) )}{( \cos(36) + \sin(36) )}$

• Proof :

I'm taking LHS to Prove this Identity -

$\Longrightarrow \sf \tan(9)$

⠀⠀⠀⠀⋆ tan θ = sin θ / cos θ

$\Longrightarrow \sf \dfrac{ \sin(9) }{ \cos(9) }$

⠀⠀⠀⠀⋆ we can write 9 = (45 - 36)

$\Longrightarrow \sf \dfrac{ \sin(45 - 36) }{ \cos(45 - 36) }$

⠀⠀⠀⠀⋆ sin(A - B) = sinAcosB - cosAsinB

⠀⠀⠀⠀⋆ cos(A - B) = cosAcosB + sinAsinB

$\Longrightarrow \sf \dfrac{ \sin(45) \cos(36) - \cos(45) \sin(36)}{ \cos(45) \cos(36) + \sin(45) \sin(36) }$

⠀⠀⠀⠀⋆ sin(45) = 1 /√2 = cos(45)

$\Longrightarrow \sf \dfrac{ \dfrac{1}{ \sqrt{2} } \times \cos(36) - \dfrac{1}{ \sqrt{2} } \times \sin(36)}{ \dfrac{1}{ \sqrt{2} } \times \cos(36) + \dfrac{1}{ \sqrt{2} } \times \sin(36) }$

$\Longrightarrow \sf \dfrac{ \cancel\dfrac{1}{ \sqrt{2} }( \cos(36) - \sin(36)) }{\cancel\dfrac{1}{ \sqrt{2} }( \cos(36) + \sin(36)) }$

$\Longrightarrow \sf \dfrac{( \cos(36) - \sin(36)) }{( \cos(36) + \sin(36)) }$

⠀

$\therefore \boxed{ \tan(9) = \dfrac{( \cos(36) - \sin(36) )}{( \cos(36) + \sin(36) )}}$