Question

Prove that?

Answer 1

See explanation.

Explanation:

`sqrt((1-sinA)/(1+sinA)) = secA -tanA`
Your first aim should be to get rid of the square root, by making everything within it squared. You'll need to use some identities for this.

`sqrt((1-sinA)/(1+sinA))`

`= sqrt( (1-sinA)/(1+sinA) *(1-sinA)/(1-sinA))`

`= sqrt( (1-sinA)^2/((1+sinA)(1-sinA)))`

Applying `(a+b)(a-b)=a^2 -b^2` for the denominator...

`= sqrt( (1-sinA)^2/(1-sin^2A) )`

Now the numerator is squared, but the denominator still needs to changed using the identity `1-sin^2A = cos^2A`

`= sqrt( (1-sinA)^2/(cos^2A))`

The squares can cancel out the square root...

`= (1-sinA)/(cosA)`

`= 1/cosA -sinA/cosA`

`= secA -tanA`

Answer 2

Kindly, refer to a Proof given in the Explanation.

Explanation:

Here is another way to solve the Problem :

We know that, `1-sin^2A=cos^2A,`

`:. (1-sinA)(1+sinA)=cosA*cosA,`

`:. (1-sinA)/cosA=cosA/(1+sinA)................(star_1).`

But, `(1-sinA)/cosA=1/cosA-sinA/cosA=secA-tanA...(star_2).`

Thus, we conclude from `(star_1) and (star_2),` that,

`secA-tanA=(1-sinA)/cosA, and, secA-tanA=cosA/(1+sinA).`

Multiplying the corresponding sides, we get,

`(secA-tanA)(secA-tanA)=(1-sinA)/cancelcosA*cancelcosA/(1+sinA),`

`i.e., (secA-tanA)^2=(1-sinA)/(1+sinA),` equivalently,

`secA-tanA=sqrt{(1-sinA)/(1+sinA)}.`

Enjoy Maths.!