Question

It is given that $a \: sinB + b \: sinB = c \:$. prove:-
$= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}$​

Answer 1

Solution:

It is given that $a \: sinB + b \: sinB = c \:$

$a \: sinB + b \: sinB = c \: (given)$

Squaring both sides,

$So, \: {(a \: sinB + b \: sinB) }^{2} = {c}^{2} .$

$= > {a}^{2} {sin}^{2} B + {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} (1 - {cos}^{2} B) + {b}^{2} (1 - {sin}^{2} B) + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} - {a}^{2} {cos}^{2} + {b}^{2}\: - {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} {cos}^{2} B - 2ab \: sinB \: cosB \: + {b}^{2} {sin}^{2} B= {a}^{2} + {b}^{2} - {c}^{2}. \:$

$= > ({a \: cosB - b \: sinB})^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}$

Hence, it is proved.

Answer 2

Given a Cos θ - b Sin θ = c (Squaring on both sides )

a2Cos2 θ + b2 Sin2 θ - 2abSin θ Cos θ = c2 ----------(1)

Let a Sin θ+ b Cos θ = k  (Squaring on both sides )

b2 Cos2 θ + a2 Sin2 θ + 2ab Sin θ Cos θ = k2 ----------(2)

Adding (1) and (2) we get

a2 + b2  = c2 + k2

k2 = a2 + b2 - c2 .

k = √(a2 + b2 - c2 )

∴ a Sin θ + b Cos θ = √(a2 + b2 - c2 )