+23 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)
+506 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.
+506 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.
