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If y(\alpha )=\sqrt{2\left [ \frac{\tan \alpha +\cot \alpha }{1+\tan ^{2}\alpha } \right ]+\frac{1}{\sin ^{2}\alpha }},\alpha \epsilon \left [ \frac{3\pi }{4},\pi \right ], then, \frac{dy}{d\alpha } at \alpha =\frac{5\pi }{6} is :
Option: 1 -\frac{1}{4}      

Option: 2 \frac{4}{3}

Option: 3 4

Option: 4 -4

 

Answers (1)

best_answer

 

 

Derivative of the Trigonometric Function (csc/sec/cot) -

Derivative of the Standard Function (Part 4)

\\\mathbf{11.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(\cot(x))=-\csc^2(x)}}

\\\mathbf{12.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(\sec(x))=\sec (x)\tan (x)}}

\\\mathbf{13.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(\csc(x))=-\csc (x)\cot (x)}}

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\\y(x)=\sqrt{2\frac{\tan x+\cot x}{1+\tan^2 x}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\frac{1}{\sin x\cos x}}{1+\frac{\sin^2 x}{\cos^2 x}}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\cos x}{\sin x}+\frac{1}{\sin^2 x}}\\\\ =\sqrt{\frac{2\cos x\sin x+1}{\sin^2 x}}\\\\ =\sqrt{\frac{(\cos x+\sin x)^2}{\sin^2 x}}\\\\ =\frac{(\cos x+\sin x)}{\sin x}

|1+\cot x|= \frac{\mathrm{d} }{\mathrm{d} x}(-1-\cot x)=\csc^2x

put x=\frac{5\pi}{6}

we get 4

Correct Option (3)

Posted by

Ritika Jonwal

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