# Question: Prove $\cot (A+ 45^{\circ}) = \frac{\csc(A)-\sec(A)}{\csc(A)+ \sec(A)}$

Question: Prove $\cot (A+ 45^{\circ}) = \dfrac{\csc(A)-\sec(A)}{\csc(A)+\sec(A)}$

My attempt:LHS

$$\cot (A+ 45^{\circ})$$

$$\frac{\cos(A+ 45^{\circ})}{\sin(A+45^{\circ})}$$

$$\cos(45^{\circ}) = \frac{1}{\sqrt2},\sin(45^{\circ}) = \frac{1}{\sqrt2}$$

$$\frac{\cos(A+ 45^{\circ})}{\sin(A+45^{\circ})}$$

$$\frac{\cos(A)\cos(45^{\circ}) - \sin(A)\sin(45^{\circ})}{\sin(A)\cos(45^{\circ}) + \cos(A)\sin(45^{\circ})}$$

$$\frac{ \frac{\cos(A)}{\sqrt2} - \frac{\sin(A)}{\sqrt2}} { \frac{\sin(A)}{\sqrt2} + \frac{\cos(A)}{\sqrt2}}$$

$$\frac{\cos(A)-\sin(A)}{\sqrt2} \times \frac{\sqrt{2}}{\sin(A)+\cos(A)}$$

$$\frac{\cos(A)-\sin(A)}{\sin(A)+\cos(A)}$$

Now I am stuck..

You are almost there. Divide the numerator and denominator of $$\frac{\cos(A) - \sin(A)}{\sin(A) + \cos(A)}$$ by $\sin(A)\cos(A)$, then simplify.
• What I meant is $$\frac{\cos(A) - \sin(A)}{\sin(A) + \cos(A)} \cdot \frac{\frac{1}{\sin(A)\cos(A)}}{\frac{1}{\sin(A)\cos(A)}}$$ Mar 16, 2016 at 2:33