Question

if secθ-cosθ=1,then tan squareθ/2=
A.√5+2
B.√5-2
C.√2-5
D.0

Answer 1

1/cos the ta - cos the ta = 1

(1 - cos^2 theta)/cos theta -1=0

1 - cos^2 theta - cos theta = 0

cos^2 theta + cos theta -1=0

cos theta = (-1 +- √1+4)/2

= (-1 +- √5 )/2

As cos theta = (1- tan^2 theta/2)/ 1 + tan^2 theta/2

= ( -1+-√5)/2 + (-1+-√5)/2 . tan^2 theta/2 -1 + tan^2 theta/2 = 0

tan^2 theta/2 ( 1+-√5)/2) =-(-3+-√5)/2

tan^2 theta/2 = -(-3+-√5)/2)/1+-√5/2

=( 3 -+ √5)/2 / 1+-√5)/2

rationalize

( 3+ √5)( 1+ √5) / 6

= 3 + 3√5 + √5 + 5)/6

= 8 + 4√5) / 6

4/3 + 2√5/3

or
( 3- √5) ( 1- √5) / -4

3 - 3√5 - √5 + 5) / -4

8 - 4√5 / -4

-2 + √5

So IT'S B