Math, asked by payal456, 9 months ago


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

Answers

Answered by MoodyCloud
13

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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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