Question

Cosec∆+ cot∆=5 then evaluate sec∆

Answer 1

it is given that,

cosec∆ + cot∆ = 5 ......(1)

we know, cosec²x - cot²x = 1

⇒(cosecx - cotx)(cosecx + cotx) = 1

⇒(cosecx - cotx) = 1/(cosecx + cotx)

so, cosec∆ - cot∆ = 1/(cosec∆ + cot∆)

⇒cosec∆ - cot∆ = 1/5 [ from equation (1), ]

now, we have to equations,

• cosec∆ + cot∆ = 5
• cosec∆ - cot∆ = 1/5

add these equations we get,

2cosec∆ = (5 + 1/5) = 26/5

or, cosec∆ = 13/5 = h/p

here, b = 5 , h = 13 so, b = √(13² - 5²) = 12

now, sec∆ = h/b = 13/12

hence, value of sec∆ = 13/12