Question

cosA-sinA=√2 then prove cosecA=2√2cosA​

Answer 1

Answer:

ANSWER

We know the formula,

sin A = 1-2cosA

so,

sin²A = 1-4cosA+4cos²A

1-cos²A = 1-4cosA+4cos²A

0 = 5cos²A-4cosA

0 = cosA(5cosA-4)

cosA=0 or cosA=4/5

If cosA=0 then one possibility is sinA=1 and in this case 2sinA-cosA=2.

At first I thought cosA=0 and sinA=-1 was possible but this is not consistent with sinA+2cosA=1.

Let's consider the case cosA=4/5.

Then sinA = ±√(1-cos²A) = ±√(1-16/25) = ±3/5

If cosA=4/5 and sinA=3/5 then sinA + 2cosA doesn't equal 1.

If cosA=4/5 and sinA=-3/5 then sinA + 2cosA does equal 1 but 2sinA-cosA= -2.

So my conclusion is that if sinA+2cosA=1 then 2sinA-cosA can equal either 2 or -2.

Answer 2

Answer:

cosA+sinA=2–√cosAcos⁡A+sin⁡A=2cos⁡A

Squaring,

cos2A+sin2A+2cosAsinA=2cos2Acos2⁡A+sin2⁡A+2cos⁡Asin⁡A=2cos2⁡A

Using cos2A=1−sin2Acos2⁡A=1−sin2⁡A

1−sin2A+sin2A+2cosAsinA=2(1−sin2A)1−sin2⁡A+sin2⁡A+2cos⁡Asin⁡A=2(1−sin2⁡A)

1+2cosAsinA=2−2sin2A1+2cos⁡Asin⁡A=2−2sin2⁡A

2cosAsinA=1−2sin2A2cos⁡Asin⁡A=1−2sin2⁡A

1−2cosAsinA=2sin2A1−2cos⁡Asin⁡A=2sin2⁡A

cos2A+sin2A−2cosAsinA=2sin2Acos2⁡A+sin2⁡A−2cos⁡Asin⁡A=2sin2⁡A

(cosA−sinA)2=2sin2A(cos⁡A−sin⁡A)2=2sin2⁡A

cosA−sinA=±2–√sinAcos⁡A−sin⁡A=±2sin⁡A

If you want answer in terms of the exact values,