Question

sinA+2 cosA = 1 prove that 2 sinA - cosA =2
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Answer 1

Answer:

2 sinA - cosA = 2

Step-by-step explanation:

= > sinA + 2 cosA = 1

= > √( sin^2 A ) + 2 cosA = 1

= > √( 1 - cos^2 A ) + 2 cosA = 1

= > √( 1 - cos^2 A ) = 1 - 2 cosA

= > 1 - cos^2 A = ( 1 - 2 cosA )^2

= > 1 - cos^2 A = 1 + 4 cos^2 A - 4 cosA

= > 5cos^2 A- 4 cosA = 0

= > cosA( 5 cosA - 4 ) = 0

= > cosA = 0 or cosA = 4 / 5

= > cosA = cos90°

= > A = 90°

Therefore :

= > cos90° = 0

= > 2 sin90° - cos90° = 2 sin90°

= > 2 sinA - cosA = 2( 1 ) { sin90° = 1 & here, 90° = A }

= > 2 sinA - cosA = 2

Hence proved.

Answer 2

$\huge\underline\mathrm{Question-}$

If $\sf{SinA+2CosA=1}$ then prove that, $\sf{2SinA-CosA=2}$

$\huge\underline\mathrm{Solution-}$

Given : $\sf{SinA+2CosA=1}$

Squaring both sides,

: $\implies$ $\sf{(SinA+2CosA)^2=(1)^2}$

By using (a+b)² = a² + b² + 2ab

: $\implies$ $\sf{Sin^2\:A\:+4Cos^2\:A\:+4SinACosA=1}$

: $\implies$ $\sf{4Cos^2\:A+4SinACosA=1-Sin^2\:A}$

By using 1 - Sin²A = Cos²A

: $\implies$ $\sf{4Cos^2\:A+4SinACosA=Cos^2\:A}$

: $\implies$ $\sf{4Cos^2\:A\:-Cos^2\:A\:+4SinACosA=0}$

: $\implies$ $\sf{3Cos^2\:A\:+4SinACosA=0}$ ________(1)

Now, let's square (2SinA - CosA),

: $\implies$ $\sf{(2SinA-CosA)^2=4Sin^2\:A\:+Cos^2\:A-4SinACosA}$

Using (1)

: $\implies$ $\sf{(2SinA-CosA)^2=4Sin^2\:A\:+Cos^2\:A+3Cos^2\:A}$

: $\implies$ $\sf{(2SinA-CosA)^2=4Sin^2\:A\:+4Cos^2\:A}$

: $\implies$ $\sf{(2SinA-CosA)^2=4(Sin^2\:A+Cos^2\:A}$

Using (Sin²A+Cos²A=1),

: $\implies$ $\sf{(2SinA-CosA)^2=4(1)}$

: $\implies$ $\sf{2SinA-CosA=\sqrt{4}}$

: $\implies$ $\sf{2SinA-CosA=2}$

Hence proved!