Question

Someone pls solve these 2 :(

Answer 1

$\large\bf{\pink{ \ Question \ 3 :-}}$

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$\large\rm{\green{ \lim\limits_{x \to 0} \dfrac{ e^{x} - e^{-x} - 2 \ \log(1+x)}{x \ \sin \ x} \ \bigg ( \dfrac{0}{0} \bigg ) }}$ form

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apply L'hospital's rule

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$\large\rm{\green{ \lim\limits_{x \to 0} \dfrac{e^{x} + e^{-x} - 2 \frac{2}{(1+x)}}{ x \ \cos \ x + \sin \ x} \ \bigg ( \dfrac{0}{0} \bigg ) }}$ form

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again apply L'hospital's rule

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$\large\rm{ \green{\lim\limits_{x \to 0} \dfrac{e^{x} - e^{-x} - 2 \frac{2}{(1+x)^{2}}}{ -x \ \sin \ x + 2 \ \cos \ x}}}$ form

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$\large\rm{\green{ = \dfrac{2}{2} = 1}}$

$\large\bf{\pink{ \ Question \ 4 :-}}$

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$\large\rm{ \blue{y = \sin^{2} \Bigg ( \cot^{-1} \sqrt{\dfrac{1+x}{1-x}} \Bigg ) }}$

consider x = cos θ

$\large\rm{ \blue{\therefore y = \sin^{2} \Bigg ( \cot^{-1} \sqrt{\dfrac{1+ \cos \theta}{1- \cos \theta}} \Bigg ) }}$

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$\large\rm{ \blue{= \sin^{2} \Bigg ( \cot^{-1} \sqrt{\dfrac{(1+ \cos \theta)^{2}}{1- \cos^{2} \theta}} \Bigg ) }}$

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$\large\rm{ \blue{= \sin^{2} \Bigg ( \cot^{-1} \sqrt{\dfrac{(1+ \cos \theta)^{2}}{\sin^{2} \theta}} \Bigg ) }}$

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$\large\rm{ \blue{= \sin^{2} \Bigg ( \cot^{-1} \bigg ( \dfrac{1+ \cos \theta}{ \sin \theta} \bigg ) \Bigg ) }}$

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$\large\rm{\blue{ = \sin^{2} \Bigg ( \cot^{-1} \dfrac{ 2 \ \cos^{2} \frac{\theta}{2}}{2 \ \sin \frac{\theta}{2} \times \frac{\theta}{2}} \Bigg )}}$

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$\large\rm{ \blue{= \sin^{2} ( \cot^{-1} \cot \frac{\theta}{2} ) }}$

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$\large\rm{ \blue{= \sin^{2} \bigg ( \dfrac{\theta}{2} \bigg ) }}$

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$\large\rm{ \blue{\dfrac{ dy}{ d \theta} = 2 \ \sin \bigg ( \dfrac{ \theta}{2} \bigg ) \cdot \dfrac{1}{2}}}$

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$\large\rm{ \blue{\dfrac{dy}{dx} = \dfrac{2 \ \sin ( \frac{\theta}{2} ) \times \cos ( \frac{\theta}{2} ) \cdot \frac{1}{2}}{ - \sin \theta}}}$

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$\large\rm{\blue{ = \dfrac{1}{2}}}$