Question

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Answer 1

${\large{\underline{\underline{\bf{Question:-}}}}}$

Prove that :

$\sf \: \sqrt{ \dfrac{cosec \: x + 1}{cosec \: x - 1} } = \dfrac{cos \: x}{1 - sin \: x}$

${\large{\bf{\underline{\underline{Required\:Solution:-}}}}}$

Take LHS :

$\colon \rightarrow \sf \: \sqrt{ \dfrac{cosec \: x + 1}{cosec \: x - 1} }$

• Rationalize the denominator.

$\colon \rightarrow \sf \: \sqrt{ \dfrac{(cosec \: x + 1)}{(cosec \: x - 1)} \times \dfrac{(cosec \: x + 1)}{(cosec \: x + 1)} }$

• Using a² - b² = (a+b) (a-b)

$\colon \rightarrow \sf \: \sqrt{ \dfrac{ {(cosec \: x + 1)}^{2} }{ {cosec}^{2} \: x - {1}^{2} } }$

$\colon \rightarrow \sf \: \dfrac{(cosec \: x + 1)}{ \sqrt{ {cosec}^{2} \: x - 1 } }$

• Change 1 into cosec² - cot² .

$\colon \rightarrow \sf \: \dfrac{(cosec \: x + 1)}{ \sqrt{ {cosec}^{2} \: x - ( {cosec}^{2} \: x - {cot}^{2} \: x } }$

$\colon \rightarrow \sf \: \dfrac{(cosec \: x + 1)}{ \sqrt{ \cancel {{cosec}^{2} \: x} - \cancel{ {cosec}^{2} \: x } + {cot}^{2} \: x }}$

$\colon \rightarrow \sf \: \dfrac{(cosec \: x+ 1)}{ \sqrt{ {cot}^{2} \: x } }$

$\colon \rightarrow \sf \: \dfrac{(cosec \: x + 1)}{cot \: x}$

• Change cosec into $\bf\dfrac{1}{sin}$ &
• Change cot into $\bf\dfrac{cos}{sin}$

$\colon \rightarrow \sf \: \dfrac{ \dfrac{1}{sin \: x} + 1 }{ \dfrac{cos \: x}{sin \: x} }$

• Take LCM

$\colon \rightarrow \sf \: \dfrac{ \dfrac{(1 + sin \: x)}{ \cancel{sin \: x}} }{ \dfrac{cos \: x}{ \cancel{sin \: x}} }$

$\colon \rightarrow \sf \: \dfrac{1 + sin \: x}{cos \: x}$

• Multiply the numerator and the denominator by (1-sin x)

$\colon \rightarrow \sf \: \frac{(1 + sin \: x)}{cos \: x} \times \frac{(1 - sin \: x)}{(1 - sin \: x) }$

• Using (a+b) (a-b) = a² - b²

$\colon \rightarrow \sf \: \frac{1 - {sin}^{2} \: x}{cos \: x(1 - sin \: x)}$

• Change 1 - sin² into cos²

$\colon \rightarrow \sf \: \frac{ \cancel{ {cos}^{2} \: x}}{ \cancel{cos \: x}(1 - sin \: x)}$

$\colon \rightarrow \boxed{ \red{ \bf \frac{cos \: x}{1 - sin \: x} }}$

So,

■ L.H.S = R.H.S

■ Hence proved

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Answer 2

Answer:

$\huge\mathcal{\green{Hola!}}$

Step-by-step explanation:

☆ To Prove:-

$\huge\sqrt{ \frac{cosec \: x + 1}{cosec \: x - 1} } = \frac{cos \: x}{1 - sin \: x}$

☆ Trigonometric identities:-

${sin}^{2} x + {cos}^{2} x = 1$

${sec}^{2} x - {tan}^{2} x = 1$

${cosec}^{2} x = 1 + {cot}^{2} x$

☆ Solution:-

L.H.S

• By rationalising the denominator in the L.H.S we have

$\huge \sqrt{ \frac{cosec \: x + 1}{cosec \: x - 1} \times \frac{cosec \: x + 1}{cosec \: x + 1} } = \sqrt{ \frac{ {(cosec \: x + 1)}^{2} }{ {cosec}^{2}x - 1 } }$

$\huge\mathcal{\green{Now,}}$

• we know that cot^2(x) = cosec^2(x) - 1

$\huge\mathcal{\green{Therefore,}}$

$\huge = > \sqrt{ \frac{(cosec \: x + 1) {}^{2} }{ {cot}^{2}x } }$

$\huge = > \sqrt({ \frac{ {cosec \: x + 1}^{} }{cot \: x}) {}^{2} }$

$\huge = > \frac{cosec \: x + 1}{cot \: x}$

$\huge\mathcal{\green{Now,}}$

• we know that;

$cosec \: x \: = \frac{1}{sin \: x} \: \: and \: \: cot \: x = \frac{cos \: x}{sin \: x}$

Putting the values, we have;

$\huge \frac{ \frac{1}{sin \: x} + 1 }{ \frac{cos \: x}{sin \: x} } = \frac{ \frac{1 + sin \: x}{sin \: x} }{ \frac{cos \: x}{sin \: x} }$

･ ➝ 1 + sin x/cos x

Now, by multiplying both the numerator and the denominator with

(1 - sin x)

we have;

･➝ cos x/ 1 - sin x = RHS

□ As LHS = RHS

□ Hence, proved.

$\huge\mathcal{\green{All \ the \ very \ best!}}$

$\huge\mathfrak{\red{@MissTranquil}}$

$\huge\fbox{\orange{be \ brainly}}$

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