Question

if cos ø=1/2then find the value of 2secø/1+tanø

Answer 1

If your question means:

2 sec ø / ( 1 + tan ø )

then

sec ø = 1 / cos ø = 1 / ( 1 / 2 ) = 2

2 sec ø = 2 ∙ 2 = 4

Cosine is positive in Quadrant I and Quadrant IV

in Quadrant I:

cos ø = 1 / 2 for ø = π / 3

tan ø = tan π / 3 = √ 3

1 + tan ø = 1 + √ 3 = √ 3 + 1

2 sec ø / ( 1 + tan ø ) = 4 / ( √ 3 + 1 )

Multiply numerator and denominator by √ 3 - 1

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 - 1 ) / [ ( √ 3 + 1 ) ( √ 3 - 1 ) ]

Since ( a + b ) ( a - b ) = a² - b²

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 - 1 ) / [ ( √ 3 )² - 1² ]

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 - 1 ) / ( 3 - 1 )

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 - 1 ) / 2

2 sec ø / ( 1 + tan ø ) = 2 ( √ 3 - 1 )

in Quadrant IV:

cos ø = 1 / 2 for ø = 5 π / 3

tan ø = tan 5 π / 3 = - √ 3

1 + tan ø = 1 - √ 3

2 sec ø / ( 1 + tan ø ) = 4 / ( 1 - √ 3 )

Multiply numerator and denominator by 1 + √ 3

2 sec ø / ( 1 + tan ø ) = 4 ( 1 + √ 3 ) / [ ( 1 - √ 3 ) ( 1 + √ 3 ) ]

Since ( a - b ) ( a + b ) = a² - b²

2 sec ø / ( 1 + tan ø ) = 4 ( 1 + √ 3 ) / [ 1² - ( √ 3 )² ]

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 + 1 ) / ( 1 - 3 )

2 sec ø / ( 1 + tan ø ) = 4 ( √ 3 + 1 ) / ( - 2 )

2 sec ø / ( 1 + tan ø ) = - 2 ( √ 3 + 1 )

Answer 2

Without finding the actual angle:
if cosø = 1/2, then secø = 2
x^2 + y^2 = r^2
1 + y^2 = 4
y = ± √3 , and sinø = ± √3/2,
and tanø = sinø/cosø
= ±(√3/2)/(1/2)
= ±√3

In quad I:
2secø/(1+tanø) = 4/(1 + (√3/2)/(1/2)) = 4/(1+√3)
in quad IV:
2secø/(1+tanø) = 4/(1 - (√3/2)/(1/2)) = 4/(1-√3)