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Question Number 108766 by bemath last updated on 19/Aug/20

\frac{B e M a t h}{★}

(1) f i n d t h e e q u a t i o n o f t h e t a n g e n t l i n e t o

t h e g r a p h o f t h e e q u a t i o n \sin^{- 1} (x) + \cos^{- 1} (y) = \frac{π}{2}

a t g i v e n p o i n t (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})

(2) I f f (x) = \lim_{t \to x} \frac{\sec t - \sec x}{t - x}, f i n d t h e v a l u e o f

f^{'} (\frac{π}{4})

(3) \lim_{x \to 1} \frac{\tan^{- 1} (x) - \frac{π}{4}}{x - 1}

Answered by Dwaipayan Shikari last updated on 19/Aug/20

3) \lim_{x \to 1} \frac{{t a n}^{- 1} \frac{x - 1}{1 + x}}{x - 1} = \frac{x - 1}{x - 1} . \frac{1}{1 + x} = \frac{1}{2}

Answered by john santu last updated on 19/Aug/20

\frac{⊸ J S ⊸}{♡}

(1) \frac{d}{d x} [\sin^{- 1} (x) + \cos^{- 1} (y)] = 0

\Rightarrow \frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - y^{2}}} y^{'} (x) = 0

g r a d i e n t \Rightarrow y^{'} (\frac{\sqrt{2}}{2}) = \frac{\sqrt{1 - \frac{1}{2}}}{\sqrt{1 - \frac{1}{2}}} = 1

∴ t h e e q u a t i o n o f t a n g e n t l i n e

y = 1 . (x - \frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}; y = x

(2) f (x) = \lim_{t \to x} \frac{\sec t - \sec x}{t - x}

\Rightarrow f (x) = \lim_{t \to x} \frac{\cos x - \cos t}{\cos t . \cos x (t - x)}

\Rightarrow f (x) = \lim_{t \to x} \frac{1}{\cos t . \cos x} . \lim_{t \to x} \frac{- 2 \sin (\frac{x + t}{2}) \sin (\frac{x - t}{2})}{t - x}

\Rightarrow f (x) = \frac{\sin x}{\cos^{2} x} \to f^{'} (x) = \frac{\cos^{3} x + 2 \sin^{2} x \cos x}{\cos^{4} x}

\Rightarrow f^{'} (x) = \frac{\cos^{2} x + 2 \sin^{2} x}{\cos^{3} x}

t h e r e f o r e f^{'} (\frac{π}{4}) = \frac{\frac{1}{2} + 2 . \frac{1}{2}}{\frac{1}{2 \sqrt{2}}} = \frac{3}{2} . 2 \sqrt{2} = 3 \sqrt{2}

(3) \lim_{x \to 1} \frac{\tan^{- 1} (x) - \frac{π}{4}}{x - 1} = \lim_{x \to 1} [\frac{1}{1 + x^{2}}] = \frac{1}{2}

Commented by bemath last updated on 19/Aug/20

n i c e \dots

Answered by mathmax by abdo last updated on 19/Aug/20

1) arcsinx + arcosy = \frac{π}{2} \Rightarrow arcosy = \frac{π}{2} - arcsnx \Rightarrow

y (x) = \cos (\frac{π}{2} - arcsinx) = \sin (arcsinx) = x \Rightarrow y^{^{'}} (x) = 1 \Rightarrow

y = y^{^{'}} (\frac{\sqrt{2}}{2}) (x - \frac{\sqrt{2}}{2}) + y (\frac{\sqrt{2}}{2}) = x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = x \Rightarrow y = x is the

equation of tangente

Answered by mathmax by abdo last updated on 19/Aug/20

2) f (x) = \lim_{t \to x} \frac{\frac{1}{sint} - \frac{1}{sinx}}{t - x} \Rightarrow f (x) = \frac{d}{dx} (\frac{1}{sinx}) = - \frac{cosx}{\sin^{2} x} \Rightarrow

f^{^{'}} (x) = - \frac{- {sinxsin}^{2} x - 2 sinx cosx cosx}{\sin^{4} x}

= \frac{\sin^{2} x + {2 \cos}^{2} x}{\sin^{3} x} = \frac{1 + \cos^{2} x}{\sin^{3} x} \Rightarrow f^{^{'}} (\frac{π}{4}) = \frac{1 + {(\frac{1}{\sqrt{2}})}^{2}}{{(\frac{1}{\sqrt{2}})}^{3}} = \frac{3}{2 (\frac{1}{2 \sqrt{2}})} = 3 \sqrt{2}

Answered by mathmax by abdo last updated on 19/Aug/20

3) \lim_{x \to 1} \frac{arctanx - \frac{π}{4}}{x - 1} = \lim_{x \to 1} \frac{\arctan (x) - \arctan (1)}{x - 1}

= \arctan^{^{'}} (1) = \frac{1}{2}