Answer:
Proof below
Step-by-step explanation:
Trigonometric Identities
Prove that:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=2+\sec x\csc x[/tex]
We need to use the following basic identities:
[tex]\sin^2x+\cos^2x=1\qquad\qquad [1][/tex]
[tex]\displaystyle \sec x=\frac{1}{\cos x}\qquad\qquad [2][/tex]
[tex]\displaystyle \csc x=\frac{1}{\sin x}\qquad\qquad [3][/tex]
Operating on the left side:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=\frac{\sin^2 x+2\cos x\sin x+\cos^2x}{\sin x\cos x}[/tex]
Applying [1]:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=\frac{1+2\cos x\sin x}{\sin x\cos x}[/tex]
Separating fractions:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}+\frac{2\cos x\sin x}{\sin x\cos x}[/tex]
Simplifying the second term:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}+2[/tex]
Applying [2] and [3]
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=\csc x\sec x+2[/tex]
Rearranging:
[tex]\displaystyle \frac{(\sin x+\cos x)^2}{\sin x\cos x}=2+\sec x\csc x[/tex]
Hence proved.
