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Question Number 118969 by benjo_mathlover last updated on 21/Oct/20

G i v e n f (x) = \frac{\sin x + \cos x}{\sin x - \cos x}

f i n d t h e v a l u e o f

f^{″} (x) + f^{'} (x) + 1 .

Answered by TANMAY PANACEA last updated on 21/Oct/20

f (x) = \frac{t a n x + 1}{t a n x - 1} = - t a n (\frac{π}{4} + x)

f^{^{'}} (x) = - {s e c}^{2} (\frac{π}{4} + x)

f^{^{″}} (x) = - 2 {s e c}^{2} (\frac{π}{4} + x) t a n (\frac{π}{4} + x)

1 + f^{^{'}} (x) + f^{^{″}} (x) = 1 - 2 {s e c}^{2} (\frac{π}{4} + x) t a n (\frac{π}{4} + x) - {s e c}^{2} (\frac{π}{4} + x)

Answered by bramlexs22 last updated on 21/Oct/20

f^{'} (x) = \frac{(\cos x - \sin x) (\sin x - \cos x) - (\sin x + \cos x) (\cos x + \sin x)}{1 - \sin 2 x}

= \frac{\sin 2 x - 1 - (1 + \sin 2 x)}{1 - \sin 2 x} = \frac{- 2}{1 - \sin 2 x}

f^{″} (x) = \frac{0 - (- 2) (- 2 \cos 2 x)}{{(1 - \sin 2 x)}^{2}} = \frac{- 4 \cos 2 x}{{(1 - \sin 2 x)}^{2}}

f^{″} (x) + f^{'} (x) + 1 = \frac{- 4 \cos 2 x - 2 (1 - \sin 2 x) + {(1 - \sin 2 x)}^{2}}{{(1 - \sin 2 x)}^{2}}

= \frac{- 4 \cos 2 x + 2 \sin 2 x - 2 + 1 - 2 \sin 2 x + \sin^{2} 2 x}{{(1 - \sin 2 x)}^{2}}

= \frac{\sin^{2} 2 x - 4 \cos 2 x - 1}{{(1 - \sin 2 x)}^{2}} = \frac{1 - \cos^{2} 2 x - 4 \cos 2 x - 1}{{(1 - \sin 2 x)}^{2}}

= \frac{- \cos 2 x (\cos 2 x + 4)}{{(1 - \sin 2 x)}^{2}}

Answered by MJS_new last updated on 21/Oct/20

\sin x = \frac{\tan x}{\sqrt{1 + \tan^{2} x}} \land \cos x = \frac{1}{\sqrt{1 + \tan^{2} x}}

let t = \tan x \Rightarrow t^{'} = 1 + t^{2}

\Rightarrow

f (x) = \frac{t + 1}{t - 1}

f^{'} (x) = - \frac{2 t^{'}}{{(t - 1)}^{2}} = - \frac{2 (t^{2} + 1)}{{(t - 1)}^{2}}

f^{″} (x) = \frac{4 t^{'} (t + 1)}{{(t - 1)}^{3}} = \frac{4 (t + 1) (t^{2} + 1)}{{(t - 1)}^{3}}

1 + f^{'} (x) + f^{″} (x) = \frac{(t + 1) (3 t^{2} + 5)}{{(t - 1)}^{3}} =

= - \frac{(1 + \tan x) (5 + {3 \tan}^{2} x)}{{(1 - \tan x)}^{3}} =

= \frac{(\sin x + \cos x) (3 + {2 \cos}^{2} x)}{{(\sin x - \cos x)}^{3}}

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