Tan3-1-tan3-1-3 then the general value of - is-

The correct option is D nπ/3+7π/36 (tan3θ> 1/tan3θ>+1)=(√(3)/1) use>componendo>and>dividendo>,>we>get> (tan3θ> ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

tan3θ-1/tan3θ...

Question

$\frac{t a n 3 θ - 1}{t a n 3 θ + 1} = \sqrt{3}$ then the general value of $θ$ is-

\frac{n π}{3} - \frac{π}{12}

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n π + \frac{7 π}{12}

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\frac{n π}{3} + \frac{7 π}{36}

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n π + \frac{π}{12}

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Solution

The correct option is D $\frac{n π}{3} + \frac{7 π}{36}$
$(\frac{t a n 3 θ - 1}{t a n 3 θ + 1}) = (\frac{\sqrt{3}}{1}) u s e c o m p o n e n d o a n d d i v i d e n d o, w e g e t (\frac{t a n 3 θ - 1 + t a n 3 θ + 1}{t a n 3 θ - 1 - t a n 3 θ - 1}) = (\frac{\sqrt{3} + 1}{\sqrt{3} - 1}) o r t a n 3 θ = (\frac{\sqrt{3} + 1}{1 - \sqrt{3}}) o r t a n 3 θ = (\frac{t a n (\frac{π}{3}) + t a n \frac{π}{4}}{1 - t a n (\frac{π}{3}) (\frac{π}{4})}) ∴ 3 θ = (\frac{π}{3}) + (\frac{π}{4}) ∴ θ = (\frac{n π}{3}) + (\frac{7 π}{36})$

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tan3θ−1tan3θ+1=√3 then the general value of θ is-

The correct option is D nπ3+7π36(tan3θ−1tan3θ+1)=(√31)usecomponendoanddividendo,weget(tan3θ−1+tan3θ+1tan3θ−1−tan3θ−1)=(√3+1√3−1)ortan3θ=(√3+11−√3)ortan3θ=(tan(π3)+tanπ41−tan(π3)(π4))∴3θ=(π3)+(π4)∴θ=(nπ3)+(7π36)

$\frac{t a n 3 θ - 1}{t a n 3 θ + 1} = \sqrt{3}$ then the general value of $θ$ is-