Question

If sec θ - tan θ; y = cosec θ + cot θ then

A. x = ( y + 1 ) / ( y - 1 )
B. x = ( y - 1 ) / ( y + 1 )
C. y = ( 1 + x ) / ( 1 - x )
D. xy + x - y + 1 = 0

Answer 1

Answer:

Answer:–

xy+x-y+1=0

Given:–

X=Secθ-tanθ

Y=Cosecθ+Cotθ

Explanation:–

It is given that,

$x = secθ - tanθ$

$x = \frac{1 - \sinθ}{cosθ}$

$y = cosecθ + cotθ$

$y = \frac{1 + cosθ}{sinθ}$

Now,

$xy + x - y + 1 = 0$

Take LHS,

➟ $xy + x - y + 1$

➟ $( \frac{1 - sinθ}{cosθ} )( \frac{1 + cosθ}{sinθ} ) + 1 + \frac{1 - sinθ}{cosθ} - ( \frac{1 + cosθ}{sinθ} )$

➟ $\frac{1 - sinθ + cosθ - sinθ.cosθ + sinθ.cosθ}{sinθ.cosθ} + \frac{1 - sinθ}{cosθ} - ( \frac{1 + cosθ}{sinθ})$

Since,

Sin²θ+Cos²θ=1

➟ $\frac{ {sin}^{2}θ + {cos}^{2}θ - sinθ + cos θ }{sinθ.cosθ} + (secθ - tanθ) - (cosecθ + cotθ)$

➟ $\frac{ {sin}^{2}θ + {cos}^{2} θ }{sinθ.cosθ} - ( \frac{sinθ - cosθ}{sinθ.cosθ} ) + secθ - tanθ - cosecθ - cotθ$

➟ $(tanθ + cotθ) - (secθ - cosecθ) + secθ - tanθ - cosecθ - cotθ$

➟ $0$

So,

$xy + x - y + 1 = 0$

Now,

↠ x=y-1/y+1

↠ xy+x=y-1

↠ xy+x-y+1=0

And also,

↠ y=1+x/1-x

↠ y-xy=1+x

↠ xy+x-y+1=0

So, The correct option are B, C and D.

Answer 2

❥ Question :

If sec θ - tan θ; y = cosec θ + cot θ then

A. x = ( y + 1 ) / ( y - 1 )

B. x = ( y - 1 ) / ( y + 1 )

C. y = ( 1 + x ) / ( 1 - x )

D. xy + x - y + 1 = 0

❥ Answer :

                                                                               

Step - 1 :

$\bf{\text { Given } x=\sec \theta-\tan \theta=\dfrac{1-\sin \theta}{\cos \theta}{$

$\bf{y={cosec} \theta+\cot \theta=\dfrac{1+\cos \theta}{\sin \theta}}$

$\bf{\text { xy }+1=\left(\dfrac{1-\sin \theta}{\cos \theta}\right)\left(\dfrac{1+\cos \theta}{\sin \theta}\right)+1=\dfrac{1-\sin \theta+\cos \theta}{\sin \theta \cos \theta}}$

$\bf{=\dfrac{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)}{\sin \theta \cos \theta}-\dfrac{(\sin \theta-\cos \theta)}{\sin \theta \cos \theta}}$

$\bf{=(tan \theta+\cot \theta)-(\sec \theta-{cosec} \theta)}$

$\bf{=({cosec} \theta+\cot \theta)-(\sec \theta-\tan \theta)=y-x }$

⇒ xy + 1= y −x

                                                                               

Step - 2 :

→ Solve for x in xy + 1= y −x

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$xy+1-1=y-x-1$

$\Rightarrow xy=y-x-1$

$\mathrm{Add\:}x\mathrm{\:to\:both\:sides}$

$xy+x=y-x-1+x$

$\Rightarrow xy+x=y-1$

$\mathrm{Factor}\:xy+x:$

$\mathrm{Factor\:out\:common\:term\:}x$

$=x\left(y+1\right)$

$x\left(y+1\right)=y-1$

$\mathrm{Divide\:both\:sides\:by\:}y+1;\quad \:y\ne \:-1$

$\dfrac{x\left(y+1\right)}{y+1}=\dfrac{y}{y+1}-\dfrac{1}{y+1};\quad \:y\ne \:-1$

$\bf{x=\dfrac{y-1}{y+1};\quad \:y\ne \:-1}$

$\star$ Hence $\bf{x=\dfrac{y-1}{y+1}$  is correct $\star$

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