Question

Heya Brainliacs!


If SinA + 2CosA = 1, then prove that -
2SinA - CosA = 2



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Answer 1

Given that

Sin A + 2 cos A = 1

Squaring on both sides, we get

(sin A + 2 cos A)^2 = 1

We know that (a+b)^2 = a^2 + b^2 + 2ab.

(sin^2 A + 4 cos^2 A + 4 sin A cos A) = 1

4 cos^2 A + 4 sin A cos A = 1 - sin^2 A

4 cos^2 A + 4 sin A cos A = cos^2 A

3 cos^2 A + 4 sin A cos A = 0    

3 cos^2 A = - 4 sin A cos A  ---- (1).

LHS = 2 sin A - cos A 

On squaring , we get

(2 sin A - cos A)^2 = 4 sin^2 A + cos^2 A - 4 sin A cos A

= 4 sin^2A + cos^2A + 3 cos^2A from (i)

= 4 sin^2 A + 4 cos^2 A

= 4(sin^2 A + cos^2 A)

= 4.

Now, (2 sin A - cos A)^2 = 4

= 2 sin A - cos A = 2

LHS  = RHS.

HENCE PROVED.

Answer 2

Heya ,

Question:-

If SinA +2CosA =1 then prove that 2SinA - CosA =2

Answer-:

SinA + 2Cos =1

squaring on both sides , we get

(SinA + 2CosA )^ 2 = 1

we know that (a+b)^2 = a^2 + b^2 +2ab

=>(sin^2A +4 cos^2A +4sin A. CosA)=1

4 cos ^2A+4 sinA cosA= 1 -sin^2A

4 cos^ 2A+4sinA cosA = cos ^2A

3cos^2A +4sinA cosA =0

3cos ^2A= - 4sinA cosA --- (1)

Given 2 sinA cosA

squaring on both sides , we get

(2sinA - cosA ) ^ 2= 4 sin ^2A + cos ^2A - 4 sinA cosA = 4 sin^2A + cos ^2A + 3cos^2A

2sinA - cos A =2

LHS= RHS

thank you:-)

Hope this helps:-)