Question

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

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

$\huge\tt{AnsweR}$

Squaring both sides:

$\sf ={(a \: sinB + b \: sinB) }^{2} = {c}^{2}$

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

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

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

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

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

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

Therefore Proved!