Question

if sin a=1/3 evaluate cos a cosec a+tan a sec a

Answer 1

sin a = 1/3
BC/AC = 1/3
BC = 1
AC = 3
By pythagoras property, 
                                    (AC)₂ = (AB)₂ + (BC)₂
                                     [3]² = (AB)₂ + (1)₂
                                     (AB)₂ = 9-1
                                               =8
                                      AB=√8=2√2
cosA cosecA + tanA secA= 2√2/3×3 + 1/2√2×3/2√2
                                           =16√2 + 3 /8

Answer 2

Trignometric Ratios

→ Sin = Perpendicular/Hypotenuse = 1/3

Hence, Perpendicular = 1 and Hypotenuse = 3.

→ H² = B² + P²

→ 3² = B² + 1²

→ √8 or 2√2 = B

• Cosec ∅ = 1/Sin ∅
• Sec ∅ = 1/Cos ∅
• Tan ∅ = Sin/Cos ∅
• Cos ∅ = Base/Hypotenuse

→ Cos A = 2√2/3

→ Sec A = 3/2√2

→ Cosec A = 3

→ Tan A = 1/2√2

[By filling respective values]

→ Cos A Cosec A + Tan A Sec A

→ 2√2/3 × 3 + (1/2√2 × 3/2√2)

→ 2√2 + 3/8

→ (16√2 + 3)/8

Answer : (16√2 + 3)/8