Question

If Sin0=4/5 then find the value of 3sec0-5Cos0​

Answer 1

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5 sec x-3 tan x = 5

or 5 sec x - 5 = 3 tan x

or 5(sec x - 1 ) = 3 tan x

or 5(1-cos x)/cos x =3.sin x/cos x

or 5(1-cos x) = 3.sin x

or 5/3=sin x/(1-cos x)

or 5/3=sin x.(1+cos x)/(1-cos x) (1+cos x)

or 5/3 = sin x.(1+cos x)/(1-cos^2x)

or 5/3=sin x(1+cos x)/sin^2x=(1+cos x)/sin x

or 5 sin x=3+3cos x

On dividing both sides by cos x .

or 5tan x =3sec x + 3

or 5tan x - 3 sec x = 3 , Answer.

Second - Method :-

Formula :-

5sec x -3.tan x = 5. …………….(1). , let 5.tan x -3.sec x= k. ……………………(2).

Squaring both sides the eqn. (1) and eqn.(2) and subtracting.

25.sec^2 x -30.sec x.tan x+9.tan^2 x -25.tan^2 x + 30.sec x.tan x - 9.sec^2 x = 25-k^2.

or, 16.(sec^2 x- tan^2 x) = 25 - k^2.

or, 16.(1) = 25 - k^2.

or, k^2 =25 - 16.

or, k^2 = 9. => k = √9 = 3. , Answer.