If-1-n-sin-2-1-n-cos-2-1-n-find-tan-2-

Question Number 21453 by Joel577 last updated on 24/Sep/17

If

(1 + n) \sin 2 θ + (1 - n) \cos 2 θ = 1 + n

find \tan 2 θ

Answered by $@ty@m last updated on 24/Sep/17

(1 - n) \cos 2 θ = (1 + n) (1 - \sin 2 θ)

\frac{1 - n}{1 + n} = \frac{1 - \sin 2 θ}{\cos 2 θ} = \frac{{(\cos θ - \sin θ)}^{2}}{\cos^{2} θ - \sin^{2} θ}

\frac{1 - n}{1 + n} = \frac{\cos θ - \sin θ}{\cos θ + \sin θ}

\frac{- 2 n}{2} = \frac{- 2 \sin θ}{2 \cos θ}, b y c o m p o n e n d o & d i v i d e n d o

\tan θ = n

∴ \tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}

\Rightarrow \tan 2 θ = \frac{2 n}{1 - n^{2}} A n s .

Answered by sma3l2996 last updated on 24/Sep/17

(1 + n) s i n 2 θ + (1 - n) c o s 2 θ = 1 + n

\Leftrightarrow (1 + n) t a n 2 θ + 1 - n = \frac{1 + n}{c o s 2 θ}

w e h a v e \frac{1}{c o s a} = \sqrt{1 + {t a n}^{2} a}

s o (1 + n) t a n 2 θ + 1 - n = (1 + n) \sqrt{1 + {t a n}^{2} 2 θ}

{(1 + n)}^{2} {t a n}^{2} 2 θ + {(1 - n)}^{2} + 2 (1 + n) (1 - n) t a n 2 θ = {(1 + n)}^{2} (1 + {t a n}^{2} 2 θ)

{(1 - n)}^{2} + 2 (1 - n^{2}) t a n 2 θ = {(1 + n)}^{2}

2 (1 - n^{2}) t a n 2 θ = {(1 + n)}^{2} - {(1 - n)}^{2} = 4 n

t a n 2 θ = \frac{2 n}{1 - n^{2}}

Answered by $@ty@m last updated on 24/Sep/17

A n o t h e r s o l u t i o n :

(1 + n) \frac{2 \tan θ}{1 + \tan^{2} θ} + (1 - n) \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = 1 + n

2 (1 + n) y + (1 - n) (1 - y^{2}) = (1 + n) (1 + y^{2}), w h e r e y = \tan θ

2 (1 + n) y + 1 - n - y^{2} + {n y}^{2} = 1 + n + y^{2} + {n y}^{2}

2 (1 + n) y = 2 (n + y^{2})

y^{2} - (1 + n) y + n = 0

y = \frac{1 + n \pm \sqrt{{(1 + n)}^{2} - 4 n}}{2}

y = \frac{1 + n \pm (1 - n)}{2}

y = 1, n

\tan 2 θ = \infty, \frac{2 n}{1 - n^{2}}

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