Question

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

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{ 1 + { \sin}^{2} x = 3 \sin x \cos x\: }$

TO PROVE

$\displaystyle \sf{ \tan x\: = 1 \: \: or \: \: \frac{1}{2} }$

PROOF

$\sf{ 1 + { \sin}^{2} x = 3 \sin x \cos x\: }$

$\sf{Dividing \: both \: sides \: by \: { \cos}^{2} x \: \: \: we \: get}$

$\displaystyle \: \sf{ \frac{1}{{ \cos}^{2} x} + \frac{{ \sin}^{2} x}{{ \cos}^{2} x} = 3 \: \: \frac{\sin x \cos x\: }{{ \cos}^{2} x} }$

$\implies \: \sf{ { \sec}^{2}x + { \tan}^{2} x = 3 \tan x \: }$

$\implies \: \sf{ 1 + { \tan}^{2}x + { \tan}^{2} x = 3 \tan x \: }$

$\implies \: \sf{ 1 +2 { \tan}^{2}x - 3 \tan x \: = 0}$

$\implies \: \sf{ 2 { \tan}^{2}x - 3 \tan x \: + 1= 0}$

$\implies \: \sf{ 2 { \tan}^{2}x -2 \tan x \: - \tan x + 1= 0}$

$\implies \: \sf{ 2 { \tan}x (\tan x \: - 1) - 1(\tan x - 1)= 0}$

$\implies \: \sf{ (2\tan x \: - 1) (\tan x - 1)= 0}$

$\implies \: \sf{ So \: either \: (2\tan x \: - 1) = 0 \: \: or \: \: (\tan x - 1)= 0}$

Now

$\displaystyle \: \: \sf{ \: (2\tan x \: - 1) = 0 \: \: gives \: \: \tan x = \frac{1}{2} }$

Again

$\implies \: \sf{ \: (\tan x \: - 1) = 0 \: \:gives\: \: \tan x = 1}$

Therefore

$\displaystyle \sf{ \tan x\: = 1 \: \: or \: \: \frac{1}{2} }$

Hence proved

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Answer 2

Given :  1+sin²x=3sinxcosx

To Find : prove that tanx=1 or 1/2

Solution:

Here is another way to solve other than above

1+sin²x=3sinxcosx

use 1 = sin²x + cos²x

=> sin²x + cos²x + sin²x=3sinxcosx

=> 2sin²x - 3sinxcosx  + cos²x = 0

=> 2sin²x - 2sinxcosx - sinxcosx  + cos²x = 0

=> 2sinx(sinx - cosx) - cosx(sinx - cosx) = 0

=> (2sinx - cosx)(sinx - cosx) = 0

=> 2sinx - cosx =  0 => sinx /cosx = 1/2 =>  tanx = 1/2

sinx - cosx = 0  => sinx = cosx => tanx = 1

tanx=1 or 1/2

QED

Hence proved

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