Question

If 1+ sin^2x = 3sinxcosx then prove that tanx = 1 or 1/2

Answer 1

1 + sin²x = 3 sinx cosx 
(1 + sin²x)² = 9 sin²x cos²x 
1 + sin⁴x + 2 sin²x = 9 sin²x (1 - sin²x) 
1 + sin⁴x + 2 sin²x = 9sin²x - 9sin⁴x 
10sin⁴x - 7sin²x + 1 = 0 

sin²x = [ -(-7) ± √((-7)² - 4(10)(1))] / 2(10) 
sin²x = [ 7 ± √(49 - 40)] / 20 
sin²x = [ 7 ± 3] / 20 
sin²x = 1/2 or 1/5 

sinx = ±1/√2 or ±1/√5 

tanx 
= sinx/cosx 
= sinx/√(1 - sin²x) 

= (±1/√2)/√(1 - 1/2) OR (±1/√5)/√(1 - 1/5) 
= ±1 OR (±1/√5)/√(4/5) 
= ±1 OR ±1/2 

Answer 2

Answer:

We have,

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

Dividing each term by cos²x,

We get,

$sec^2x+tan^2x=3tanx$

$1+tan^2x+tan^2x=3tanx$

( Because, sec²x = 1 + tan²x )

$1+2 tan^2x=3tanx$

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

Let tan x = t,

$2t^2-3t+1=0$

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

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

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

If 2t-1=0 ⇒ t = 1/2

If t-1 = 0 ⇒ t =1,

Thus, tan x = 1/2 or tan x = 1.

Hence, proved.