Question

$\frac{ {tan}^{2} x}{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x } = \frac{1}{ {sin}^{2}x - {cos}^{2} x}$
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Answer 1

Answer:

$\huge\star \underline{ \boxed{ \purple{Answer}}}\star$

$\sf\green{\frac{ {tan}^{2} x}{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x }} = \pink\sf{\frac{1}{ {sin}^{2}x - {cos}^{2} x}}$

Step-by-step explanation:

Given :-

$\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x }}$

To prove :-

$\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x } = \frac{1}{ {sin}^{2}x - {cos}^{2} x}}$

Solution :-

$\huge\sf\red{L.H.S}$

➝$\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x }}$

➝$\sf{\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta}{cos^{2}\theta}-1} + \frac{\frac{1}{sin^{2}\theta}}{\frac{1}{cos^{2}\theta} - \frac{1}{sin^{2}\theta}}}$

➝$\sf{\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{sin^{2}\theta cos^{2}\theta}} + \frac{\frac{1}{sin^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{sin^{2}\theta cos^{2}\theta }}}$

➝$\sf{\frac{ sin^{2}\theta cos^{2}\theta}{cos^{2}\theta(sin^{2}\theta - cos^{2}\theta)} + \frac{ sin^{2}\theta cos^{2}\theta}{sin^{2}\theta(sin^{2}\theta-cos^{2}\theta}}$

➝$\sf{\frac{ sin^{2}\theta}{sin^{2}\theta-cos^{2}\theta} + \frac{ cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}}$

➝$\sf{\frac{ sin^{2}\theta + cos^{2}\theta}{sin^{2}\theta - cos^{2}\theta}}$

➝$\sf{\frac{1}{sin^{2}\theta - cos^{2}\theta}}$

$\huge\sf\green{=\:R.H.S}$

$\huge\mathcal\blue{Hence \: proved}$

$\huge\mathfrak\pink{Hope \: it \: helps \: you}$

$\huge\mathfrak\pink{friend}$