[tex] \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} [/tex] About the author Clara
Answer: [tex] \huge\star \underline{ \boxed{ \purple{Answer}}}\star[/tex] [tex]\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}} [/tex] Step-by-step explanation: Given :– [tex]\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x – 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x – {cosec}^{2} x }}[/tex] To prove :– [tex]\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}} [/tex] Solution :– [tex]\huge\sf\red{L.H.S}[/tex] ➝[tex]\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x – 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x – {cosec}^{2} x }}[/tex] ➝[tex]\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}}}[/tex] ➝[tex]\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 }}}[/tex] ➝[tex]\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}}[/tex] ➝[tex]\sf{\frac{ sin^{2}\theta}{sin^{2}\theta-cos^{2}\theta} + \frac{ cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}}[/tex] ➝[tex]\sf{\frac{ sin^{2}\theta + cos^{2}\theta}{sin^{2}\theta – cos^{2}\theta}} [/tex] ➝[tex]\sf{\frac{1}{sin^{2}\theta – cos^{2}\theta}}[/tex] [tex]\huge\sf\green{=\:R.H.S}[/tex] [tex]\huge\mathcal\blue{Hence \: proved} [/tex] [tex]\huge\mathfrak\pink{Hope \: it \: helps \: you} [/tex] [tex]\huge\mathfrak\pink{friend} [/tex] Reply
Answer:
[tex] \huge\star \underline{ \boxed{ \purple{Answer}}}\star[/tex]
[tex]\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}} [/tex]
Step-by-step explanation:
Given :–
[tex]\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x – 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x – {cosec}^{2} x }}[/tex]
To prove :–
[tex]\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}} [/tex]
Solution :–
[tex]\huge\sf\red{L.H.S}[/tex]
➝[tex]\sf{\frac{ {tan}^{2} x}{ {tan}^{2} x – 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x – {cosec}^{2} x }}[/tex]
➝[tex]\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}}}[/tex]
➝[tex]\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 }}}[/tex]
➝[tex]\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}}[/tex]
➝[tex]\sf{\frac{ sin^{2}\theta}{sin^{2}\theta-cos^{2}\theta} + \frac{ cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}}[/tex]
➝[tex]\sf{\frac{ sin^{2}\theta + cos^{2}\theta}{sin^{2}\theta – cos^{2}\theta}} [/tex]
➝[tex]\sf{\frac{1}{sin^{2}\theta – cos^{2}\theta}}[/tex]
[tex]\huge\sf\green{=\:R.H.S}[/tex]
[tex]\huge\mathcal\blue{Hence \: proved} [/tex]
[tex]\huge\mathfrak\pink{Hope \: it \: helps \: you} [/tex]
[tex]\huge\mathfrak\pink{friend} [/tex]