Answer:
Therefore,
[tex]\dfrac{\cot^{2}x}{\csc x +1}=\dfrac{1-\sin x}{\sin x}[/tex] ....Proved
Step-by-step explanation:
To Prove:
[tex]\dfrac{\cot^{2}x}{\csc x +1}=\dfrac{1-\sin x}{\sin x}[/tex]
Proof:
Left Hand Side = [tex]\dfrac{\cot^{2}x}{\csc x +1}[/tex]
Using Identity [tex]\cot^{2}x=\csc^{2}-1[/tex]
=[tex]\dfrac{\csc^{2}x-1}{\csc x +1}[/tex]
Using identity a² - b² =( a - b )( a + b )
=[tex]\dfrac{(\csc x-1)(\csc x+1)}{\csc x +1}[/tex]
=[tex](\csc x-1)[/tex]
Now Using identity [tex]\csc x=\dfrac{1}{\sin x}[/tex] we get
=[tex](\dfrac{1}{\sin x}-1)[/tex]
=[tex](\dfrac{1-\sin x}{\sin x})[/tex]
=Right Hand Side
Therefore,
[tex]\dfrac{\cot^{2}x}{\csc x +1}=\dfrac{1-\sin x}{\sin x}[/tex] ....Proved
