Question

If sec theta + cos theta =2 find sec square theta +codex square theta

Answer 1

Answer:

Explanation:EXPLANATION(Long):

Sec(theta) + cos(theta) = 5/2

Sec(theta) + 1/sec(theta) = 5/2

(sec^2(theta)+1) / sec(theta) = 5/2

2sec^2(theta) + 2 = 5sec(theta)

2sec^2(theta) - 5sec(theta) + 2 = 0

Let sec(theta)=x

then,

2x^2 - 5x + 2 = 0

2x^2 - (4+1)x + 2 = 0

2x^2 - 4x -x + 2 = 0

2x(x-2)-1(x-2)=0

(2x-1)(x-2)=0

Either (2x-1)=0 => x=1/2 => sec(theta)=1/2 [Since sec(theta)=x]

Or (x-2)=0 => x=2 => sec(theta)=2 [Since sec(theta)=x]

Then sec(theta) = 1/2 or 2

Case 1: sec(theta) = 1/2

Now we have Sec(theta) + cos(theta) = 5/2

1/2 + cos(theta) = 5/2

cos(theta) =(5/2 - 1/2)

cos(theta)=2

then,

Sec(theta) - cos(theta) = (1/2 - 2) = -3/2

Case 2: sec(theta) = 2

Now we have Sec(theta) + cos(theta) = 5/2

2 + cos(theta) = 5/2

cos(theta) =(5/2 - 2)

cos(theta)=1/2

then,

Sec(theta) - cos(theta) = (2 - 1/2) = 3/2

therefore

sec(theta) - cos(theta)=3/2 or-3/2

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Ans2:

EXPLANATION(Shortcut):

Sec(theta) + cos(theta) = 5/2

(Sec(theta) + cos(theta))^2 = (5/2)^2

Sec(theta)^2 + cos(theta)^2 + 2*Sec(theta)*cos(theta) = 25/4

Sec(theta)^2 + cos(theta)^2 + 2*Sec(theta)*(1/Sec(theta)) = 25/4

Sec(theta)^2 + cos(theta)^2 + 2 = 25/4

[Sec(theta)^2 + cos(theta)^2 + 2]- 4 = [25/4] - 4 [Subtracting both sides by 4]

Sec(theta)^2 + cos(theta)^2 - 2 = 9/4

Sec(theta)^2 + cos(theta)^2 - 2*Sec(theta)*(1/Sec(theta)) = 9/4

Sec(theta)^2 + cos(theta)^2 - 2*Sec(theta)*cos(theta) = 9/4

(Sec(theta) - cos(theta))^2 = (3/2)^2

Sec(theta) - cos(theta) = +-3/2