Question

Of sin^(-1)x + tan^(-1)x = pi/2 then prove that 2x^(2) + 1 = 5^(1/2) plzz solve this <br />

Answer 1

We know that arcsinx+arccosx=π/2.arcsin⁡x+arccos⁡x=π/2.

In the question we have arcsinx+arctanx=π/2arcsin⁡x+arctan⁡x=π/2

So it follows that

arccosx=arctanxarccos⁡x=arctan⁡x

(Now we convert arctanx−−−−−−>arccosxarctan⁡x−−−−−−>arccos⁡x Or vice versa . Conversions like these become pretty easy when you visualize a right angled triangle .)



arccosx=arccos(1/(√1+x2))arccos⁡x=arccos⁡(1/(1+x2))

(Note that we have converted arctanx−−−−−>arccos(1/(√1+x2arctan⁡x−−−−−>arccos⁡(1/(1+x2))

Now x=1/(√1+x2)x=1/(1+x2)

Square both sides .

We get : x2=1/(1+x2)x2=1/(1+x2)

Cross multiply

x2+x4=1x2+x4=1

Rearranging , we get

x4+x2−1=0x4+x2−1=0

(Voila , a quadratic equation . However , you can solve it without invoking the Quadratic formula . Infact I explained this just a few days back : Athreya Sriram's answer to Was the 2016 isc maths inverse question wrong? )

I’ll add an excerpt from my answer :

Now we continue

x4+x2−1=0x4+x2−1=0

Now one method quite obviously is using quadratic equations . As that has been covered in all the other answers , I’ll use a different and equally simple approach .

We add and subtract the square of half of the coefficient of x2:x2:

x4+x2−1+1/4−1/4=0x4+x2−1+1/4−1/4=0

Then ,

we get :

(x4+x2+1/4)−5/4=0(x4+x2+1/4)−5/4=0

Which is

(x2+1/2)2−(5–√/2)2=0(x2+1/2)2−(5/2)2=0

Now ,

x2+1/2=5–√/2x2+1/2=5/2

Then , if we multiply 2 on both sides .

2x2+1=5–√2x2+1=5

Now In my honest opinion , I really find completing the squares a lot more elegant than quadratic equations . ( Yeah I used completing the squares in the paper too )

Bear with me . I’m still a novice . Feel free to point out any mistakes .