tan a / 1+sec a + 1+sec a /tan a = 2 csc a. prove this is true.?

Answer 1

To prove that: tan a / ( 1 + sec a ) + ( 1 + sec a ) / tan a = 2 csc a.

Replace tan a with sin a / cos a, and replace sec a with 1 / cos a.
Let s = sin a, and c = cos a. To prove:

( s / c ) / ( 1 + 1/c ) + ( 1 + 1/c ) / ( s / c ) = 2 / s

Simplify left side by multiplying num & den by c :

s / ( c + 1 ) + ( c + 1 ) / s

Add the fractions:

[ s^2 + (c+1)^2 ] / [ s ( c + 1 ) ]

= [ s^2 + c^2 + 2 c + 1 ] / [ s ( c + 1 ) ]

Using s^2 + c^2 = 1 ( sin^2 a + cos^2 a = 1 ) :

= [ 2 ( c + 1 ) ] / [ s ( c + 1 ) ] = 2 / s

Answer 2

I assume you mean tan a / (1 + sec a) + (1 + sec a)/tan a = 2 csc a.

Use tan a = sin a / cos a and sec a = 1 / cos a and you get

(sin a)/[cos a(1 + 1/cos a)] = sin a /(cos a + 1).

The second fraction is just the reciprocal of this, (cos a + 1)/sin a.

Multiply the first fraction by (cos a - 1)/(cos a - 1) and get

sin a(cos a - 1) / (cos^2 a - 1) = sin a(cos a - 1)/(-sin^2 a)
= (1 - cos a)/sin a.

The goal here was to get both fractions to have the same denominator, sin a.

Now you have

(1 - cos a)/sin a + (cos a + 1)/sin a =
(1 - cos a + cos a + 1)/sin a =
2/sin a =
2 csc a.