Math, asked by akkiabhi69, 3 months ago

please solve this question fast​

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Answered by anindyaadhikari13
7

Required Answer:-

Given:

  •   \sf \cot(x)  =  \dfrac{2mn}{ {m}^{2} -  {n}^{2}  }

To find:

  • The value of sec(x)

Solution:

We have,

 \sf \cot(x)  =  \dfrac{2mn}{ {m}^{2} -  {n}^{2}  }

 \sf \implies \cot^{2} (x)  =   \bigg(\dfrac{2mn}{ {m}^{2} -  {n}^{2}  }  \bigg)^{2}

 \sf \implies \tan^{2} (x)  =   \bigg(\dfrac{ {m}^{2}  -  {n}^{2} }{2mn}  \bigg)^{2}

We know that,

 \sf \implies 1 +  { \tan}^{2}(x) =  { \sec}^{2}(x)

 \sf \implies \sec^{2} (x)  =   \bigg(\dfrac{ {m}^{2}  -  {n}^{2} }{2mn}  \bigg)^{2}  + 1

 \sf \implies \sec^{2} (x)  =   \bigg(\dfrac{ {m}^{4} + {n}^{4}  - 2 {m}^{2} {n}^{2}  + 4 {m}^{2} {n}^{2} }{4 {m}^{2}  {n}^{2} }  \bigg)

 \sf \implies \sec^{2} (x)  =   \bigg(\dfrac{ {m}^{4} + {n}^{4}  + 2 {m}^{2} {n}^{2} }{4 {m}^{2}  {n}^{2} }  \bigg)

 \sf \implies \sec (x)  =  \sqrt{ \bigg(\dfrac{ {m}^{4} + {n}^{4}  + 2 {m}^{2} {n}^{2} }{4 {m}^{2}  {n}^{2} }  \bigg)}

 \sf \implies \sec (x)  =  \sqrt{ \bigg(\dfrac{ ({m}^{2})^{2}  +( {n}^{2})^{2} + 2  \times {m}^{2}  \times {n}^{2} }{{(2mn)}^{2}}  \bigg)}

 \sf \implies \sec (x)  =  \sqrt{ \bigg(\dfrac{ ({m}^{2} +  {n}^{2} )^{2}}{{(2mn)}^{2}}  \bigg)}

 \sf \implies \sec (x)  =  \sqrt{ \bigg(\dfrac{ {m}^{2} +  {n}^{2} }{2mn}  \bigg) ^{2} }

 \sf \implies \sec (x)  =  \dfrac{ {m}^{2} +  {n}^{2} }{2mn}

Hence, the value of sec(x) is (m² + n²)/2mn.

Answer:

  •  \sf\sec (x)  =  \dfrac{ {m}^{2} +  {n}^{2} }{2mn}
Answered by Anisha5119
4

Answer:

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