Question

How do you slove this ?

`(1-tg^2 22.5°)/(tg22.5°)`

Answer 1

I tried this:

Explanation:

I would use the fact that:

`tan(x)=sin(x)/cos(x)`

and write it as:

`(1-sin^2(22.5^@)/cos^2(22.5^@))/(sin(22.5^@)/cos(22.5^@))=`

let us rearrange it:

`(cos^2(22.5^@)-sin^2(22.5^@))/cos^cancel(2)(22.5^@)*cancel(cos(22.5^@))/sin(22.5^@)=`

now the tricky part...
Let us multiply and divide by `2`:

`2/2*(cos^2(22.5^@)-sin^2(22.5^@))/(cos(22.5^@)sin(22.5^@))=`

Let us remember some trig identities

that allows us to change top and bottom as:

`2[cos(2*22.5^@)]/sin(2*22.5^@)=2cos(45^@)/sin(45^@)=2*(sqrt(2)/2)/(sqrt(2)/2)=2*1/1=2`

Answer 2

`"I tried this after"color(blue)" Mr. Gio'"`

Only to help student with different method.

`(1-tg^2 22.5^circ)/(tg22.5^circ)=2`

Explanation:

We know that.

`color(blue)(tan^2x=(1-cos2x)/(1+cos2x)`

Take, `x=22.5^circ`

#tan^2 22.5^circ=(1-cos45^circ)/(1+cos45^circ)=(1- 1/sqrt2)/(1+1/sqrt2)=(sqrt2-1)/(sqrt2+1)#

#tan^2 22.5^circ=(sqrt2-1)/(sqrt2+1)xx(sqrt2-1)/(sqrt2-1)= ((sqrt2-1)^2)/(2-1)#

`color(red)(tan^2 22.5^circ=(sqrt2-1)^2=>tan 22.5^circ=sqrt2-1`

Now,

`(1-tg^2 22.5^circ)/(tg22.5^circ)=(1-(sqrt2-1)^2)/(sqrt2-1)`

#(1-tg^2 22.5^circ)/(tg22.5^circ)=(cancel1-2+2sqrt2- cancel1)/(sqrt2-1)=(2sqrt2-2)/(sqrt2-1)#

#(1-tg^2 22.5^circ)/(tg22.5^circ)=(2cancel((sqrt2- 1)))/cancel((sqrt2-1))=2#