Question

${ \tan }^{2} Ѳ + { \cot }^{2} Ѳ + 2 = { \sec}^{2} {cosec}^{2} Ѳ$
plss solve this give me more than 1 solution​

Answer 1

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Question :-

${ \tan }^{2} Ѳ + { \cot }^{2} Ѳ + 2 = { \sec}^{2} {cosec}^{2} Ѳ$

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Answer:-

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Step by step explanation:-

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☞ Case:- I

First take LHS side ,

$⇢ { \tan}^{2} Ѳ + { \cot }^{2} Ѳ + 2 \\ \\ ⇢(1 + { \tan}^{2} Ѳ) + (1 + { \cot }^{2}Ѳ) \\ \\ ⇢ { \sec }^{2} Ѳ + {cosec}^{2} Ѳ \\ \\ ⇢ \frac{1}{ { \cos }^{2} Ѳ} + \frac{1}{ { \sin }^{2}Ѳ } \\ \\ ⇢ \frac{ { \sin }^{2}Ѳ + { \cos }^{2}Ѳ}{ { \cos }^{2}Ѳ { \sin}^{2} Ѳ } \\ \\ ⇢ \frac{1}{ { \cos }^{2} Ѳ { \sin }^{2}Ѳ } \\ \\ ⇢ {cosec}^{2} Ѳ { \sec}^{2} Ѳ$

Hence , LHS = RHS

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☞ Case II

First take LHS side

$⇢ { \tan }^{2}Ѳ + { \cot }^{2} Ѳ + 2 \\ \\ ⇢1 + { \tan }^{2} Ѳ + { \cot }^{2} Ѳ + 1 \\ \\ ⇢(1 + { \tan }^{2} Ѳ) + { \cot}^{2} Ѳ(1 + { \tan }^{2}Ѳ) \\ \\ ⇢ (1 + { \tan }^{2} Ѳ)(1 + { \cot }^{2} Ѳ) \\ \\ ⇢ { \sec }^{2} Ѳ {cosec}^{2} Ѳ$

Hence , LHS = RHS

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Hence proved

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