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# Trig Problems

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prove these identities without interacting across the equals sign (i.e. subtracting from both sides or cross multiplyin)

1.

1/(csc x + cot x) = (1-cos x)/sin x

2.

sec x/(1+cos x) = csc^2(x) (sec x -1)

Apr 22, 2021

#1
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1.

$$\frac{1}{\csc x + \cot x}\\ =\frac{1}{\frac{1}{\sin x} + \frac{\cos x }{\sin x}}\\ =\frac{1}{\frac{1+\cos x}{\sin x}}\\ =\frac{\sin x}{1+ \cos x}\\ = \frac{(\sin x)(1- \cos x)}{(1+\cos x)(1- \cos x)}\\ = \frac{(\sin x)(1- \cos x)}{1-(\cos x)^2}$$

by the Pythagorean identity, $$(\sin x)^2 + (\cos x)^2 = 1$$ (you could prove using Pythagorean's theorem), which means that $$1-(\cos x)^2=(\sin x)^2$$. Substitute that back in:

$$\frac{(\sin x)(1-\cos x)}{(\sin x)(\sin(x)}=\frac{1-\cos x}{\sin x}$$ and we are done.

2.

$$(\csc x)^2(\sec x-1)\\ =(\frac{1}{(\sin x)^2})(\frac{1}{\cos x}-1)$$

Use the Pythagorean identity again:

$$(\frac{1}{(1+\cos x)(1-\cos x)})(\frac{1-\cos x}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\frac{1}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\sec x)\\ =\frac{\sec x}{1+\cos x}$$

and we are done.

Apr 22, 2021
edited by textot  Apr 22, 2021

#1
+2

1.

$$\frac{1}{\csc x + \cot x}\\ =\frac{1}{\frac{1}{\sin x} + \frac{\cos x }{\sin x}}\\ =\frac{1}{\frac{1+\cos x}{\sin x}}\\ =\frac{\sin x}{1+ \cos x}\\ = \frac{(\sin x)(1- \cos x)}{(1+\cos x)(1- \cos x)}\\ = \frac{(\sin x)(1- \cos x)}{1-(\cos x)^2}$$

by the Pythagorean identity, $$(\sin x)^2 + (\cos x)^2 = 1$$ (you could prove using Pythagorean's theorem), which means that $$1-(\cos x)^2=(\sin x)^2$$. Substitute that back in:

$$\frac{(\sin x)(1-\cos x)}{(\sin x)(\sin(x)}=\frac{1-\cos x}{\sin x}$$ and we are done.

2.

$$(\csc x)^2(\sec x-1)\\ =(\frac{1}{(\sin x)^2})(\frac{1}{\cos x}-1)$$

Use the Pythagorean identity again:

$$(\frac{1}{(1+\cos x)(1-\cos x)})(\frac{1-\cos x}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\frac{1}{\cos x})\\ =(\frac{1}{(1+\cos x)})(\sec x)\\ =\frac{\sec x}{1+\cos x}$$

and we are done.

textot Apr 22, 2021
edited by textot  Apr 22, 2021