Question

Question #c3e29

Answer 1

Given #csc A - cot A = 1/x.. .(1)#

Now
`cscA+cot A=(csc^2A-cot^2A)/(cscA+cotA)`

`=>cscA+cot A=x......(2)`

Adding (1) and (2) we get

`2cscx=x+1/x`

`=>cscx=1/2(x+1/x)=1/2(x^2+1)/x`

Subtracting (1) from (2) we get

`2cotA=x-1/x`

`cotA=1/2(x-1/x)=1/2(x^2-1)/x`

Now
`sec A =cscA/cotA= (x^2 + 1)/ (x^2 - 1)`

Answer 2

Please see below.

Explanation:

Let `cscA-cotA=1/x`.......[1]

We know that,

`rarrcsc^2A-cot^2A=1`

`rarr(cscA-cotA)*(cscA+cotA)=1`

`rarr1/x(cscA+cotA)=1`

`rarrcscA+cotA=x`....[2]

Adding equations [1] and [2],

`rarrcscA-cotA+cscA+cotA=1/x+x`

`rarr2cscA=(x^2+1)/x`.....[3]

Subracting equation [1] from [2],

`rarrcscA+cotA-(cscA-cotA)=x-1/x`

`rarrcscA+cotA-cscA+cotA=(x^2-1)/x`

`rarr2cotA=(x^2-1)/x`.......[4]

Dividing equation [3] by [4],

`rarr(2cscA)/(2cotA)=((x^2+1)/x)/((x^2-1)/x)`

`rarr(1/sinA)/(cosA/sinA)=(x^2+1)/(x^2-1)`

`rarrsecA=(x^2+1)/(x^2-1)` Proved...

Regards to dk_ch sir