Question

what is the answer
please 1,2

Answer 1

1. LHS = $\frac{cos9+sin9}{cos9-sin9}$
            = $\frac{cot9+1}{cot9-1}$ [divided by sin9]
            = $\frac{cot9cot45+1}{cot9-cot45}$
            = $cot(45-9) = cot36$

2. LHS = $\frac{sin3x}{sin x} - \frac{cos3x}{cosx}$
            = $\frac{sin3x cosx-cos3x sin x }{sin x cos x }$
            = $\frac{sin(3x-x)}{sin x cos x}$
            = $\frac{sin(2x)}{sin x cos x} = \frac{2 sin x cos x}{sin x cos x} = 2$

Answer 2

{ \pink{\boxed{ \red{\bf{\therefore \; \; x = \dfrac{5}{3} \; \; OR\; \; 2 }}}}}[/