Prove that?

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2 Answers
Aug 20, 2017

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

Aug 20, 2017

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.!