# 1+sec(theta)-tan(theta)/ 1+sec(theta)+tan(theta) =and given options are  1-cos(theta)/sin(theta) 1-sin(theta)/cos(theta) 1+cos(theta)/sin(theta) 1+sin(theta)/cos(theta)

Lab Bhattacharjee
121 Points
9 years ago
$\sec^2y-\tan^2y=1 \implies\dfrac{1+\sec y-\tan y}{1+\sec y+\tan y} =\dfrac{\sec^2y-\tan^2y+\sec y-\tan y}{1+\sec y+\tan y} =\sec y-\tan y=\dfrac{1-\sin y}{\cos y}$
Kelvin
11 Points
6 years ago
Stupid no clear explanations Dont ever give shprt answers And make a fool like u even foolersWorst fellows...
faizan7401
11 Points
6 years ago
Since We Know That Sec²x - Tan²x = 1Then (Secx - Tanx)(Secx + Tanx) = 1{ From (a²-b²) = (a+b)(a-b) }Then Secx - Tanx = 1/(Secx + Tanx)→The Eq. (1 + Secx - Tanx)/(1 + Secx + Tanx) Can be written as (1 + 1/(Secx + Tanx))/(1 + Secx + Tanx)Now Solving Numerator : 1 + 1/(Secx + Tanx) = (Secx + Tanx + 1)/(Secx + Tanx)Now Shifting (Secx + Tanx) To Denominator : (Secx + Tanx + 1)/((Secx + Tanx)× (Secx + Tanx + 1))Secx + Tanx + 1 gets Cancelled.....!!!!!→1/(Secx + Tanx) = Secx - Tanx. (Refer First Three Steps if you didn`t get this :) )→1/Cosx - Sinx/Cosx Since Denominators Are same (Cosx) We can Add the Numerators→(1 - Sinx)/Cosx. .......!!!!!Which Is Your Opt. 2..!!!!:)
Rishi Sharma
4 years ago
Dear Student,

Since We Know That
Sec²x - Tan²x = 1
Then (Secx - Tanx)(Secx + Tanx) = 1 { From (a²-b²) = (a+b)(a-b) }
Then Secx - Tanx = 1/(Secx + Tanx)
→The Eq. (1 + Secx - Tanx)/(1 + Secx + Tanx)
Can be written as (1 + 1/(Secx + Tanx))/(1 + Secx + Tanx)
Now Solving Numerator :
1 + 1/(Secx + Tanx) = (Secx + Tanx + 1)/(Secx + Tanx)
Now Shifting (Secx + Tanx) To Denominator :
(Secx + Tanx + 1)/((Secx + Tanx)× (Secx + Tanx + 1))
Secx + Tanx + 1 gets Cancelled
→1/(Secx + Tanx) = Secx - Tanx.
→1/Cosx - Sinx/Cosx
Since Denominators Are same (Cosx)