Answer:
The correct answer is:
Option: A
A. [tex]\dfrac{\sec t+\csc t}{1+\cot t}=\sec t[/tex]
Step-by-step explanation:
A)
[tex]\dfrac{\sec t+\csc t}{1+\cot t}=\sec t\\\\\\i.e.\\\\\\\dfrac{\dfrac{1}{\cos t}+\dfrac{1}{\sin t}}{1+\dfrac{\cos t}{\sin t}}=\sec t\\\\\\\dfrac{\dfrac{\cos t+\sin t}{\cos t\sin t}}{\dfrac{\sin t+\cos t}{\sin t}}=\sec t\\\\\\\dfrac{\sin t}{\sin t\cos t}=\sec t\\\\\\\dfrac{1}{\cos t}=\sec t\\\\\\\\\sec t=\sec t[/tex]
B)
[tex]\dfrac{\sec t-\csc t}{1-\cot t}=\cos t\\\\\\i.e.\\\\\\\dfrac{\dfrac{1}{\cos t}-\dfrac{1}{\sin t}}{1-\dfrac{\cos t}{\sin t}}=\cos t\\\\\\\dfrac{\dfrac{\sin t-\cos t}{\cos t\sin t}}{\dfrac{\cos t-\sin t}{\sin t}}=\cos t\\\\\\\dfrac{-\sin t}{\sin t\cos t}=\cos t\\\\\\\dfrac{-1}{\cos t}=\cos t\\\\\\\\\-sec t=\cos t[/tex]
This is a false relation.
Since, we know that:
[tex]\sec t=\dfrac{1}{\cos t}[/tex]
C)
[tex]\dfrac{\cos t-\sin t}{1-\cot t}=\cos t[/tex]
i.e.
[tex]\dfrac{\cos t-\sin t}{1-\dfrac{\cos t}{\sin t}}=\cos t\\\\\\\dfrac{\cos t-\sin t}{\dfrac{\sin t-\cos t}{\sin t}}=\cos t\\\\\\-\dfrac{1}{\sin t}=\cos t[/tex]
which is again a wrong identity
D)
[tex]\dfrac{\sin (t+2\pi)}{\cos (\pi+t)}=\cos t[/tex]
We know that:
[tex]\sin (2\pi+t)=\sin t\\\\and\\\\\cos (\pi+t)=-\cos t[/tex]
i.e.
[tex]\dfrac{\sin t}{-\cos t}=\cos t[/tex]
i.e.
[tex]-\csc t=\cos t[/tex]
which is again a wrong identity.
