Question

If `Sin17^@ =x/y` , then show that `Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))` ?

If `Sin17^@ =x/y` , then show that `Sec17^@ - sin73^@ = x^2/(ysqrt((y^2 - x^2))`

Answer 1

`LHS=sec17^@-sin73^@`

`=1/(cos17^@)-sin(90^@-17^@)`

`=1/(cos17^@)-cos17^@`

`=(1-cos^2 17^@)/cos17^@`

`=(sin^2 17^@)/sqrt(1-sin^2 17^@)`

`=(x^2/y^2)/sqrt(1-x^2/y^2)`

`=x^2/y^2xxy/sqrt(y^2-x^2)`

`=x^2/(ysqrt(y^2-x^2))=RHS`

Answer 2

See full explanation for solution.

Explanation:

By the cofunction identity:
`sin(73^circ)=cos(90^circ-73^circ)=cos(17^circ)`

So we have `sec(17^circ)-cos(17^circ)`. We know that `sec(x)=1/cos(x)` so we can rewrite this expression:

`1/cos(17^circ)-cos(17^circ)=(1-cos^2(17^circ))/cos(17^circ)`

By the Pythagorean ID, `sin^2(x)+cos^2(x)=1` and so `1-cos^2(x)=sin^2(x)`:

`sin^2(17^circ)/cos(17^circ)`

Rearranging `sin^2(x)+cos^2(x)=1` we can also see that `cos(x)=sqrt(1-sin^2(x))`.

(Since `17^circ` is in QI all trig functions are positive.)

`sin^2(17^circ)/sqrt(1-sin^2(17^circ))`.

Now we substitute `x/y` for `sin(17^circ)`:

`(x/y)^2/sqrt(1-(x/y)^2)`.

Now we do some algebra:

`(x^2/y^2)/sqrt(1-x^2/y^2)=(x^2/y^2)/sqrt((y^2-x^2)/y^2)`

`=(x^2/y^2)/((1/y)sqrt(y^2-x^2))=(y*(x^2/y^2))/sqrt(y^2-x^2)`

`=(x^2/y)/sqrt(y^2-x^2)=x^2/(y*sqrt(y^2-x^2))`, as required.