Question

1+√3tan^2(x)=(1+√3)tanx then find x​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf1 + \sqrt{3} \tan {}^{2} x = (1 + \sqrt{3} ) \tan x$

$\;\;\underline{\textbf{\textsf{ To Find:-}}}$

• Value of x

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\\ \underline{\textsf{Given that - }}$

$\sf1 + \sqrt{3} \tan {}^{2} x = (1 + \sqrt{3} ) \tan x$

$\\ \underline{\textsf{Then:- }}$$\\ \\ \sf \dashrightarrow \sqrt{3} \tan {}^{ 2} x - (1 + \sqrt{3} ) \tan x + 1 = 0 \\ \\ \sf\dashrightarrow \sqrt{3} \tan {}^{2}x - \tan x - \sqrt{3} \tan x + 1 = 0 \\ \\ \sf \dashrightarrow \sqrt{3} \tan {}^{2}x - \sqrt{3} \tan x - \tan x + 1 = 0 \\ \\ \sf\dashrightarrow \sqrt{3} \tan x( \tan x - 1) - 1( \tan x - 1) = 0 \\ \\ \sf \dashrightarrow ( \sqrt{3} \tan x - 1)( \tan x - 1) = 0$

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$\\ \underline{\textsf{One term: - }}$

$\dashrightarrow \sf\sqrt{3} \tan x - 1 = 0 \\ \\ \sf \dashrightarrow \tan x = \frac{1}{ \sqrt{3} } \\ \\ \dashrightarrow \sf \tan x = \tan30 \degree \\ \\ \sf\dashrightarrow x = 30\degree \\ \\ \dashrightarrow \sf x = \dfrac{ \pi}{6}$

$\\ \underline{\textsf{Another term:- }}$

$\sf\dashrightarrow \tan x - 1 = 0 \\\\ \sf \dashrightarrow tan x = 1 \\\\ \sf \dashrightarrow \tan x = \tan 45\degree \\\\ \sf \dashrightarrow x = 45\degree \\\\ \dashrightarrow \sf x = \dfrac{\pi}{4}$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Value of x is \textbf{ 45°(π/4) or 30°(π/6) }}}.$

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Answer 2

Answer:

π/6 is the answer

Step-by-step explanation:

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