Question

Question #2144a

Answer 1

It is proved below:

Explanation:

`tan a+sec a=x/y`
`or,sin a/cos a+1/cos a=x/y`
`or,(sin a+1)/cos a=x/y`
`or,(sin a+1)^2/cos^2 a=x^2/y^2`
`or,(sina+1)^2/(1-sin^2a)=x^2/y^2`
`or,(sina+1)^2/((1+sin a)(1-sina))=x^2/y^2`
`or,(sina+1)/(1-sina)=x^2/y^2`
`or,(sina+1-1+sina)/(sina+1+1-sina)=(x^2-y^2)/(x^2+y^2)" "`[By division-addition method]
`or,(2sina)/(2)=(x^2-y^2)/(x^2+y^2)`
`so,sina=(x^2-y^2)/(x^2+y^2)color(blue)([Proved.])`

Answer 2

Given `sec(a)+tan(a)=x/y......(1)`

We know

`sec^2(a)-tan^2(a)=1`

`=>(sec(a)+tan(a))(sec(a)-tan(a))=1`

`=>sec(a)-sec(a)=1/(sec(a)+tan(a))`

`=>sec(a)-tan(a)=1/(x/y)`

`=>sec(a)-tan(a)=y/x.......(2)`

Adding (1) and (2) we get

`=>2sec(a)=x/y+y/x=(x^2+y^2)/(xy)`

`=>sec(a)=(x^2+y^2)/(2xy)....(3)`

Now subtracting (2) from (1) we get

`=>2tan(a)=x/y-y/x=(x^2-y^2)/(xy)`

`=>tan(a)=(x^2-y^2)/(2xy) .....(4)`

Dividing (4) by (3) we have

`tan(a)/sec(a)=(x^2-y^2)/(x^2+y^2)`

`=>sin(a)=(x^2-y^2)/(x^2+y^2)`