Question

prove that cos9+sin9/cos9-sin9=cot36

Answer 1

LHS = (Cos9° + Sin9°) / (Cos9° - Sin9°) 
Divide by Nr and Dr  by  Cos9° ...
LHS = (1 + Tan9°) / (1 - Tan9°)
        = (Tan 45° + Tan 9°) / (1 - Tan 45° * Tan 9°) 
        = Tan (45°+9°) 
        = Tan 54°
        = Cot 36°.
Hope it helps
And thanks for asking doubt

Answer 2

Answer and Explanation:

To prove : $\frac{\cos 9+\sin 9}{\cos 9-\sin 9}=\cot 36$

Proof :

Take LHS,

$\frac{\cos 9+\sin 9}{\cos 9-\sin 9}$

Take cos 9 common,

$=\frac{1+\frac{\sin 9}{\cos 9}}{1-\frac{\sin 9}{\cos 9}}$

$=\frac{1+\tan 9}{1-\tan 9}$

$=\frac{\tan 45+\tan 9}{1-\tan 9\tan 45}$

We know formula,

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Here, A=45 and B=9

$=\tan(45+9)$

$=\tan(54)$

$=\tan(90-36)$

We know, $\tan (90-\theta)=\cot \theta$

$=\cot 36$

$=RHS$

Hence proved.