Question

if x=seca+tana and y=coseca+cota then prove that dy/dx=-(1+y^2)/(1+x^2)
solve this with steps

Answer 1

Given,

$\longrightarrow x=\sec a+\tan a$

Differentiating wrt $a,$

$\longrightarrow \dfrac{dx}{da}=\dfrac{d}{da}(\sec a+\tan a)$

$\longrightarrow \dfrac{dx}{da}=\sec a\tan a+\sec^2 a$

$\longrightarrow \dfrac{dx}{da}=\sec a(\sec a+\tan a)$

$\longrightarrow \dfrac{dx}{da}=x\sec a\quad\quad\dots(1)$

Given,

$\longrightarrow y=\csc a+\cot a$

Differentiating wrt $a,$

$\longrightarrow \dfrac{dy}{da}=\dfrac{d}{da}(\csc a+\cot a)$

$\longrightarrow \dfrac{dy}{da}=-\csc a\cot a-\csc^2 a$

$\longrightarrow \dfrac{dy}{da}=-\csc a(\csc a+\cot a)$

$\longrightarrow \dfrac{dy}{da}=-y\csc a\quad\quad\dots(2)$

Dividing (2) by (1),

$\longrightarrow \dfrac{dy}{da}\div\dfrac{dx}{da}=-\dfrac{y\csc a}{x\sec a}$

$\longrightarrow \dfrac{dy}{dx}=-\dfrac{y\csc a}{x\sec a}\quad\quad\dots(3)$

We see that,

$\longrightarrow x^2=(\sec a+\tan a)^2$

$\longrightarrow x^2=\sec^2 a+\tan^2 a+2\sec a\tan a$

Adding 1,

$\longrightarrow 1+x^2=\sec^2 a+\tan^2 a+1+2\sec a\tan a$

Since $\sec^2a-\tan^2a=1,$

$\longrightarrow 1+x^2=\sec^2 a+\tan^2 a+\sec^2a-\tan^2a+2\sec a\tan a$

$\longrightarrow 1+x^2=2\sec^2 a+2\sec a\tan a$

$\longrightarrow 1+x^2=2\sec a(\sec a+\tan a)$

$\longrightarrow 1+x^2=2x\sec a$

$\longrightarrow x\sec a=\dfrac{1+x^2}{2}\quad\quad\dots(4)$

Similarly,

$\longrightarrow y^2=(\csc a+\cot a)^2$

$\longrightarrow y^2=\csc^2 a+\cot^2 a+2\csc a\cot a$

Adding 1,

$\longrightarrow 1+y^2=\csc^2 a+\cot^2 a+1+2\csc a\cot a$

Since $\csc^2a-\cot^2a=1,$

$\longrightarrow 1+y^2=\csc^2 a+\cot^2 a+\csc^2a-\cot^2a+2\csc a\cot a$

$\longrightarrow 1+y^2=2\csc^2 a+2\csc a\cot a$

$\longrightarrow 1+y^2=2\csc a(\csc a+\cot a)$

$\longrightarrow 1+y^2=2y\csc a$

$\longrightarrow y\csc a=\dfrac{1+y^2}{2}\quad\quad\dots(5)$

Putting (4) and (5) in (3) we get,

$\longrightarrow \dfrac{dy}{dx}=-\dfrac{\left(\dfrac{1+y^2}{2}\right)}{\left(\dfrac{1+x^2}{2}\right)}$

$\longrightarrow\underline{\underline{\dfrac{dy}{dx}=-\dfrac{1+y^2}{1+x^2}}}$

Hence Proved!

Answer 2

Answer:

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