Question

If cosx+3sinx=5 then prove that tanx@+secx=2

Answer 1

(Sec x)^2 - (tan x)^2 = 1

Given 2 - (cos x)^2 = 3*sin x * cos x

Divide both sides by (cos x)^2

2 /(cos x)^2 - 1 = 3 * sinx/ cos x

2 (Sec x)^2 - 1 = 3 tan x

If tan x = t then (Sec x)^2 = 1+t^2

Therefore we have

2 + 2*t^2 - 1 = 3t

2t^2 - 3t + 1 = 0

2t^2 - 2t - t + 1 = 0

(2t-1)(t-1) = 0

So t = 1 or 1/2

Since the question says sin x is not equal to cos x then tan x automatically becomes 1/2.

Answer 2

2-cos^2 x=3sin x cos x

1+1-cos^2 x=3sin x cos x

1+sin^2 x=3 sin x cos x

1+sin^2 x+cos^2 x=3sinx cos x+cos^2 x ( adding cos^2 x both sides)

2=cos x(3 sin x +cos x)

2/cos x=3 sin x+cos x

2 sec x=3 sin x+ cos x

2 sec^2 x=3 sin x sec x+ cos x sec x

2 sec^2 x= 3sin x/cos x+1

2 sec^2 x=3 tan x+ sec^2 x-tan^2 x

( By writing 1=sec^2 x-tan^2 x)

Sec^2 x+tan^2 x=3tan x

sec^2 x-tan^2 x=3 tan x-2tan^2 x

1=3 tan x-2 tan^2 x

2 tan^2 x-3 tan x+1=0

By solving it we are getting two values of tan x that are 1/2 and 1

But the value of 1 will be rejected because tan^-1 (1)=45° and for 45° both sin x = tan x and it is provided in the question that the value of sin x is not equal to cos x hence the value of tan x =1/2=0.5