Question

* यदि cote =
7/24
है तो
sin Ø.secØ/
1+ cos ec Ø.tanØ
का मान ज्ञात करें।​

Answer 1

Answer:-

(Theta is taken as "A").

Given:

cot A = 7/24

We know that,

cot ∅ = Adjacent side/Opposite side

Hence,

Adjacent side/Opposite side = 7/24

Using Pythagoras Theorem,

→ (Hypotenuse)² = (Opposite side)² + ( Adjacent side)²

→ (Hypotenuse)² = (7)² + (24)²

→ Hypotenuse = √(49 + 576)

→ Hypotenuse = √625

→ Hypotenuse = 25

Hence,

sin A = Opposite side/Hypotenuse

→ sin A = 7/25

sec A = Hypotenuse/Adjacent side

→ sec A = 25/24

We have To find:

sin A * sec A/(1 + Cosec A * tan A)

Using , Cosec A = 1/sin A and tan A = sin A/Cos A in denominator we get,

→ sin A * sec A / [1 + (1/sin A)*(sin A/Cos A)]

→ sin A * sec A/(1 + 1/Cos A)

Writing 1/Cos A as sec A in denominator we get,

→ sin A * sec A/(1 + sec A)

Putting the values of sin A , sec A we get,

→ [ (7/25) * (25/24) ] / [ 1 + 25/24 ]

→ (7/24) / [ (24 + 25)/24) ]

→ 7/24 * 24/49

→ 1/7

Hence, the value of (sin A * sec A)/(1 + Cosec A * tan A) is 1/7.