Question

secx+tanx=3 then sinx??? ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

tan x + sec x = 4

Or, tan x = 4 - sec x

Or, (tan x)^2 = ( 4- sec x )^2

Or, (tan x)^2 = 4^2 - 2*4*sec x + (sec x)^2

Or, (tan x)^2 = 16 - 8sec x + (sec x)^2

Or, (tan x)^2 - (sec x)^2 = 16 - 8sec x

Or, -1 = 16 - 8sec x

Or, 8sec x = 16+1 = 17

Or, sec x = 17/8

Or, cos x = 8/17 [as sec x = 1/cos x]

Or, (cos x)^2 = (8/17)^2 = 64/289

Or, (cos x)^2 - 1 = 64/289 - 1 = - 225/289

[ as (cos x)^2 + (sin x)^2=1 ]

So, (cos x)^2 - [ (cos x)^2 + (sin x)^2 ] = - 225/289

Or, (cos x)^2 - (cos x)^2 - (sin x)^2 = - 225/289

Or, -(sin x)^2 = - 225/289

Or, (sin x)^2 = 225/289

Or, sin x = √(225/289)

Or, sin x = 15/17 (Ans.)