sec-sec-sin-2-2-3-sin-1-has-the-roots-are-1-and-2-Find-the-value-of-tan-1-tan-2-

Question Number 115345 by bemath last updated on 25/Sep/20

\sec θ (\sec θ (\sin^{2} θ) + 2 \sqrt{3} \sin θ) = 1

h a s t h e r o o t s a r e θ_{1} a n d θ_{2} . F i n d t h e

v a l u e o f \tan θ_{1} \times \tan θ_{2} .

Answered by bobhans last updated on 25/Sep/20

\Leftrightarrow \frac{\sin^{2} θ}{\cos θ} + 2 \sqrt{3} \sin θ = \cos θ

\Leftrightarrow 2 \sqrt{3} \sin θ . \cos θ = \cos 2 θ

\Rightarrow \sqrt{3} \sin 2 θ = \cos 2 θ \to \tan 2 θ = \frac{1}{\sqrt{3}}

\to \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{1}{\sqrt{3}} \Rightarrow \tan^{2} θ + 2 \sqrt{3} \tan θ - 1 = 0

j u s t a p p l y {V i e t a}^{'} s r u l e w e g e t

\tan θ_{1} \times \tan θ_{2} = - 1

Answered by Dwaipayan Shikari last updated on 25/Sep/20

\frac{\sin^{2} θ}{\cos^{2} θ} + \frac{2 \sqrt{3} \sin θ}{\cos θ} = 1

\tan^{2} θ + 2 \sqrt{3} \tan θ - 1 = 0

\tan θ = \frac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2} = 2 - \sqrt{3} or - (2 + \sqrt{3})

\tan θ_{1} = 2 - \sqrt{3}

\tan θ_{2} = - (2 + \sqrt{3})

\tan θ_{1} . \tan θ_{2} = - 1

Answered by MJS_new last updated on 25/Sep/20

\frac{1}{c} (\frac{s^{2}}{c} + 2 \sqrt{3} s) = 1

t^{2} + 2 \sqrt{3} t - 1 = 0

\Rightarrow t_{1} \times t_{2} = - 1

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