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

Question

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..

N. F. Taussig · Accepted Answer · 2016-03-16 02:30:45Z

2

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.

answered Mar 16, 2016 at 2:30

N. F. Taussig

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$\begingroup$ I always get confused by the saying "Divide numerator and denominator" by "so and so" could you explain it to me? $\endgroup$
– bigfocalchord
Mar 16, 2016 at 2:31
$\begingroup$ What I meant is $$\frac{\cos(A) - \sin(A)}{\sin(A) + \cos(A)} \cdot \frac{\frac{1}{\sin(A)\cos(A)}}{\frac{1}{\sin(A)\cos(A)}}$$ $\endgroup$
– N. F. Taussig
Mar 16, 2016 at 2:33
$\begingroup$ We're you able to complete the proof? $\endgroup$
– N. F. Taussig
Mar 16, 2016 at 12:11
$\begingroup$ Yes I was , thanks for the help! $\endgroup$
– bigfocalchord
Mar 16, 2016 at 18:01

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Kaynex · Accepted Answer · 2016-03-16 02:39:14Z

0

Start working with the right side. Convert all trig functions to sin and cos, cross multiply the numerator and denominator.

answered Mar 16, 2016 at 2:39

Kaynex

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2 Answers 2