Show that tan(21sin−143)=34−7 and justify why the other value 34+7 is isgnored?

To prove :tan(21sin−143)=34−7Proof :Let 21sin−143=θ⇒sin−143=2θ⇒sin2θ=43⇒1+tan2θ2tanθ=43⇒3+3tan2θ=8tanθ⇒3tan2θ−8tanθ+3=0Let tanθ=y∴3y2+8y+3=0⇒y=2×3+8±64−4×3×3=68±28⇒y=232[4±7]⇒tanθ=34±7Since, −2π≤sin−143≤2π⇒−4π≤21sin−143≤4π⇒−4π≤θ≤4π∴tan(4−π)≤tanθ≤tan4π⇒−1≤tanθ≤1Ignoring 34+7 as 34+7>1, We getθ=tan−1[34±7]∴LHS=tan[21sin−1(43)]=tantan−1(34−7)=34−7=RHS

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Show that ({1}{2}^{-1}{3}{4})={4-{7}}{3} and justify why the

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Show that $tan (\frac{1}{2} sin^{- 1} \frac{3}{4}) = \frac{4 - 7}{3}$ and justify why the other value $\frac{4 + 7}{3}$ is isgnored?

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Show that

tan (\frac{1}{2} sin^{- 1} \frac{3}{4}) = \frac{4 - 7}{3}

and justify why the other value

\frac{4 + 7}{3}

is isgnored?

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

To prove :

tan (\frac{1}{2} sin^{- 1} \frac{3}{4}) = \frac{4 - 7}{3}

Proof :

Let

\frac{1}{2} sin^{- 1} \frac{3}{4} = θ \Rightarrow sin^{- 1} \frac{3}{4} = 2 θ

\Rightarrow sin 2 θ = \frac{3}{4} \Rightarrow \frac{2 tan θ}{1 + tan ^{2} θ} = \frac{3}{4}

\Rightarrow 3 + 3 tan^{2} θ = 8 tan θ

\Rightarrow 3 tan^{2} θ - 8 tan θ + 3 = 0

Let

tan θ = y

∴ 3 y^{2} + 8 y + 3 = 0

\Rightarrow y = \frac{+ 8 \pm 64 - 4 \times 3 \times 3}{2 \times 3} = \frac{8 \pm 28}{6}

\Rightarrow y = \frac{2 [ 4 \pm 7 ]}{23}

\Rightarrow tan θ = \frac{4 \pm 7}{3}

Since,

- \frac{π}{2} \leq sin^{- 1} \frac{3}{4} \leq \frac{π}{2}

\Rightarrow - \frac{π}{4} \leq \frac{1}{2} sin^{- 1} \frac{3}{4} \leq \frac{π}{4}

\Rightarrow - \frac{π}{4} \leq θ \leq \frac{π}{4}

∴ tan (\frac{- π}{4}) \leq tan θ \leq tan \frac{π}{4}

\Rightarrow - 1 \leq tan θ \leq 1

Ignoring

\frac{4 + 7}{3}

\frac{4 + 7}{3} > 1

, We get

θ = tan^{- 1} [\frac{4 \pm 7}{3}]

∴ L H S = tan [\frac{1}{2} sin^{- 1} (\frac{3}{4})]

= tan tan^{- 1} (\frac{4 - 7}{3}) = \frac{4 - 7}{3} = R H S

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Question Text	Show that $tan (\frac{1}{2} sin^{- 1} \frac{3}{4}) = \frac{4 - 7}{3}$ and justify why the other value $\frac{4 + 7}{3}$ is isgnored?
Answer Type	Text solution:1
Upvotes	150

Show that tan(21​sin−143​)=34−7​​ and justify why the other value 34+7​​ is isgnored?

Text solutionVerified

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Show that $tan (\frac{1}{2} sin^{- 1} \frac{3}{4}) = \frac{4 - 7}{3}$ and justify why the other value $\frac{4 + 7}{3}$ is isgnored?