SSC CHSL 2025 PYQ — Question 54

Has been given :

sin A = 3/5

cos B = 5/13

According to the formula:

sin(A + B) = sin A cos B + cos A sin B

sin²θ + cos²θ = 1

cos A = √(1 − sin²A)

= √(1 − (3/5)²)

= √(1 − 9/25)

= √(16/25)

= 4/5

sin B = √(1 − cos²B)

= √(1 − (5/13)²)

= √(1 − 25/169)

= √(144/169)

= 12/13

sin A cos B + cos A sin B$= \left(\frac{3}{5} \times \frac{5}{13}\right) + \left(\frac{4}{5} \times \frac{12}{13}\right) = \frac{63}{65}$