Question

if cosecA-cotA=1/x, then prove that cosA=x^2-1/x^2+1​

Answer 1

Step-by-step explanation:

We know that cosec²A-cot²A=1

So (cosec-cot)(cosec+cot)=1. (a+b)(a-b)=a²-b²

1/x×(cosec+cot)=1

cosec+cot=x

Adding Value of cosec-cot and cosec+cot

2cosec=x²+1/x

sin=2x/x²+1

We also know that

sin²+cos²=1

So cos²=1-4x2/(x²+1)²

cos²=(x²+1)²-4x²/(x²+1)²

cos²=x⁴+1+2x²-4x²/(x²+1)²

cos²=x⁴-2x²+1/(x²+1)²

cos²=(x²-1)²/(x²+1)²

cos=x²-1/x²+1

Hence proved

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Answer 2

$: \longrightarrow{ \underline{{ \huge{ \mathfrak{A}}} \mathfrak{nswer :-}}}$

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Given :

$\mathrm{ \: cosecA \: - cot A\: = \frac{1}{x} \: \: \: ...(i)}$

To prove :

$\mathrm{cosA = \frac{ {x}^{2} - 1 }{ {x}^{2} + 1 } }$

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We know that,

$\longrightarrow{ \mathrm{cosec²A - cot²A = 1}}$

since,

a² - b² = (a+b) (a-b)

${ \mathrm{ \longrightarrow\: (cosecA + cotA) \: (cosecA - cotA) = 1}}$

$\mathrm{ \longrightarrow{ \: (cosecA + cotA) \: ( \dfrac{1}{x}) = 1 }}$

$\mathrm{ \longrightarrow{ \:cosecA + cotA = x \: \: \: ...(ii) }}$

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On adding eq. (i) and (ii),

$\mathrm{ \longrightarrow{ \:cosec A - \cancel{cotA}+ cosecA + \cancel{cotA} = \frac{1}{x} + x }}$

$\mathrm{ \longrightarrow{ \:2 \: cosec A = \dfrac{ {x}^{2} + 1 }{x} }}$

$\mathrm{ \longrightarrow{ \: cosecA = \dfrac{ {x}^{2} + 1 }{2x} }}$

We know that,

cosecA = 1/sinA

$\mathrm{ \longrightarrow{ \: \dfrac{1}{sinA} = \dfrac{ {x}^{2} + 1}{2x} }}$

Now, cross multiply it,

$\mathrm{ \longrightarrow{ \: \dfrac{2x}{ {x}^{2} + 1} = sinA}}$

We got the value of sinA.

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We know that,

$\mathrm{ \longrightarrow{ \: {sin}^{2} A + {cos}^{2} A = 1 }}$

$\mathrm{ \longrightarrow{ \: {cos}^{2}A = 1 - {sin}^{2} A }}$

$\mathrm{ \longrightarrow{ \: {cos}^{2}A = 1 - {\huge{(}} \dfrac{2x}{ {x}^{2} - 1 } { \huge{)}} ^{2} }}$

$\mathrm{ \longrightarrow{ \: {cos}^{2} A = 1 - \dfrac{4 {x}^{2} }{( {x}^{2} + 1) ^{2} } }}$

By taking L.C.M ,

$\mathrm{ \longrightarrow{ \: {cos}^{2} A = \dfrac{( {x}^{2} + 1) ^{2} - 4 {x}^{2} }{ {( {x}^{2} + 1)}^{2} } }}$

We know that,

(a + b)² = a² + b² + 2ab

$\mathrm{ \longrightarrow{ \: {cos}^{2} A \: = \dfrac{ ({x}^{4} + 1 + 2 {x}^{2} )- 4 {x}^{2} }{ {( {x}^{2} } + 1)^{2} } }}$

$\mathrm{ \longrightarrow{ \: {cos}^{2} A \: = \dfrac{ {x}^{4} - 2 {x}^{2} + 1 }{ {( {x}^{2} + 1)}^{2} } }}$

By,

(a - b)² = a² + b² - 2ab

$\mathrm{ \longrightarrow{ \: {cos}^{2} A = \dfrac{ {( {x}^{2} - 1)}^{2} }{ {( {x}^{2} + 1)}^{2} } }}$

$\mathrm{ \longrightarrow{ \: {cos} A = \sqrt{ \dfrac{ {( {x}^{2} - 1)}^{2} }{ {( {x}^{2} + 1) }^{ 2} } } }}$

Root gets cancel with the squares,

$\red{\mathrm{ \longrightarrow{ \: {cos}A = \dfrac{ {x}^{2} - 1 }{ {x}^{2} + 1 } }}}$

___________proved____

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