Question

15. If (sec A-tan A) = x then prove that
(1 + x^2)/(1-x^2)=cosec A.​

Answer 1

split Sec^2A and Tan^2A using identities in each case

1. 1+x^2
2. 1-x^2

Then put the value in LHS

Answer 2

The given question is If (sec A-tan A) = x then prove that

$(1 + x^2)/(1-x^2)=cosec A.$

It is given that sec A - tan A =x.

we have to prove

$\frac{1 + {x}^{2} }{ 1 - {x}^{2} } = \csc(a)$

sec A - tan A =x

which implies

x= sec A - tan A.

On squaring both sides, we get

${x}^{2} = {( \sec(a) - \tan(a) ) }^{2}$

we can use the algebraic formula here,

The algebraic formula is

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

${( \ sec(a) - \tan(a) ) }^{2} = { \sec}^{2} a \: + { \tan }^{2} a - 2 \sec( a ) \tan(a)$

In the above equation apply the componendo and dividendo.

$\frac{ {x}^{2} + 1}{ {x}^{2} - 1} = \frac{ { \sec }^{2} a + { \tan }^{2} a - 2 \sec( a) \tan(a) + 1 }{ { \sec }^{2} a + { \tan }^{2} a - 2 \sec(a) \tan(a) - 1 }$

This is the expansion of an algebraic formula.

${ \tan}^{2} a \: + 1 = { \sec}^{2} a$

${ \sec}^{2} a - 1 = { \tan}^{2} a$

let us substitute these values in the above division.

$= \frac{ { \sec }^{2}a + ( { \tan }^{2}a + 1) - 2 \sec(a) \tan(a) }{( { \sec}^{2} a- 1) + { \tan}^{2} a- 2 \sec(a) \tan(a) }$

$= \frac{ { \sec }^{2} a + { \sec}^{2} a - 2 \sec(a) \tan(a) }{ { \sec }^{2}a + { \tan}^{2} a- 2 \sec(a) \tan(a) }$

we can combine the like terms

$= \frac{2 { \sec}^{2}a - 2 \sec(a) \tan(a) }{2 { \tan}^{2}a -2 \sec(a) \tan(a) } \\ \frac{2 \sec(a)(( \sec(a) - \tan(a)) }{2 \tan(a) (( \sec(a) - \tan(a) }$

On proceeding, we get the value as

$\sec(a) - \tan(a) \\ in \: the \: numerator \: and \: denominator \: gets \: cancelled$

therefore, the final answer is

$\frac{ \sec(a) }{ \tan(a) } = \csc(a)$

$\frac{1}{ \cos(a) } \times \frac{ \cos(a) }{ \sin(a) } = \frac{1}{ \sin(a) } = \csc(a)$

Hence proved

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