Question

a cosA + b sinB = c prove that a sinA - b cosA = root a2+ b2 - c2



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Answer 1

Answer:

Step-by-step explanation:

acosA + bsinB = c  (1)

asinB - bcosA = k (2)

We need to find k

Lets square both the equations

$(1)^{2}$  ---->

$(acosA + bsinB)^{2} = c^{2}$

$a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinB = c^{2}$  (3)

$(2)^{2}$  ---->

$(asinB - bcosA)^{2} = k^{2}$

$a^{2}sin^{2}B+b^{2}cos^{2}A-2absinBcosA = k^{2}$  (4)

Add (3) and (4)

$a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinB$+$a^{2}sin^{2}A+b^{2}cos^{2}B-2absinBcosA$ = $c^{2}+k^{2}$

$a^{2}[cos^{2}A+sin^{2}A] +b^{2}[sin^{2}B+cos^{2}B]+2abcosAsinB-2absinBcosA = c^{2}+k^{2}$

$a^{2}+b^{2}=c^{2}+k^{2}$

$k = \sqrt{a^{2}+b^{2}-c^{2}}$

$asinB-bcosA = \sqrt{a^{2}+b^{2}-c^{2}}$

Answer 2

Answer:

acosA + bsinB = c (1)

asinB - bcosA = k (2)

We need to find k

Lets square both the equations

(1)^{2}(1)

2

---->

(acosA + bsinB)^{2} = c^{2}(acosA+bsinB)

2

=c

2

a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinB = c^{2}a

2

cos

2

A+b

2

sin

2

B+2abcosAsinB=c

2

(3)

(2)^{2}(2)

2

---->

(asinB - bcosA)^{2} = k^{2}(asinB−bcosA)

2

=k

2

a^{2}sin^{2}B+b^{2}cos^{2}A-2absinBcosA = k^{2}a

2

sin

2

B+b

2

cos

2

A−2absinBcosA=k

2

(4)

Add (3) and (4)

a^{2} cos^{2}A+b^{2}sin^{2}B+2abcosAsinBa

2

cos

2

A+b

2

sin

2

B+2abcosAsinB +a^{2}sin^{2}A+b^{2}cos^{2}B-2absinBcosAa

2

sin

2

A+b

2

cos

2

B−2absinBcosA = c^{2}+k^{2}c

2

+k

2

a^{2}[cos^{2}A+sin^{2}A] +b^{2}[sin^{2}B+cos^{2}B]+2abcosAsinB-2absinBcosA = c^{2}+k^{2}a

2

[cos

2

A+sin

2

A]+b

2

[sin

2

B+cos

2

B]+2abcosAsinB−2absinBcosA=c

2

+k

2

a^{2}+b^{2}=c^{2}+k^{2}a

2

+b

2

=c

2

+k

2

k = \sqrt{a^{2}+b^{2}-c^{2}}k=

a

2

+b

2

−c

2

asinB-bcosA = \sqrt{a^{2}+b^{2}-c^{2}}asinB−bcosA=

a

2

+b

2

−c

2

is ur ans...