Question

prove that cos 9° + sin 9° / cos 9° - sin 9° = cot 36°​

Answer 1

Answer:

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

cos9−sin9

cos9+sin9

=cot36

Proof :

Take LHS,

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

cos9−sin9

cos9+sin9

Take cos 9 common,

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

1−

cos9

sin9

1+

cos9

sin9

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

1−tan9

1+tan9

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

1−tan9tan45

tan45+tan9

We know formula,

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

1−tanAtanB

tanA+tanB

Here, A=45 and B=9

=\tan(45+9)=tan(45+9)

=\tan(54)=tan(54)

=\tan(90-36)=tan(90−36)

We know, \tan (90-\theta)=\cot \thetatan(90−θ)=cotθ

=\cot 36=cot36

=RHS=RHS

Answer 2

Answer:

By this u understand properly