Correct Answer  Option 1 : M
^{1}L
^{1}T
^{1}, M
^{1}L
^{2}T
^{2}
CONCEPT:
Dimensions:

Dimensions of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.
EXPLANATION:
Given F = X cos(Pt) + Y sin(Qs), t = time and s = distance
The dimension of F, t and s is given as,
⇒ [F] = [M^{1}L^{1}T^{2}]
⇒ [t] = [M0L0T1]
⇒ [s] = [M0L1T0]
 Since the trigonometric function is a dimensionless quantity. So, Pt and Qs is also a dimensionless quantity.
So,
⇒ [P][t] = [M0L0T^{0}]
⇒ [P][M0L0T1] = [M0L0T0]
⇒ [P] = [M0L0T^{1}] (1)
⇒ [Q][s] = [M0L0T0]
⇒ [Q][M0L1T0] = [M0L0T0]
⇒ [Q] = [M0L^{1}T0] (2)
 The dimension of the F, X and Y must be equal.
⇒ [F] = [X] = [Y] = [M1L1T2] (3)
By equation 1 and equation 3,
\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\frac{\left [ M^{0}L^{0}T^{1} \right ]}{\left [ M^{1}L^{1}T^{2} \right ]}\)
\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{1}L^{1}T^{1} \right ]\)
By equation 1 and equation 3,
\(\Rightarrow \frac{\left [ Q \right ]}{\left [ Y \right ]}=\frac{\left [ M^{0}L^{1}T^{0} \right ]}{\left [ M^{1}L^{1}T^{2} \right ]}\)
\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{1}L^{2}T^{2} \right ]\)
 Hence, option 1 is correct.