Question

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

Actually your question has some mistakes.

Its should be =>

QUESTION

a Sin B + b Cos B = c

-Prove that

( a CosB - b SinB )= √( a² + b² - c²)

ANSWER

Here we have =>

a sin B + b Cos B = c

=>(a Sin B + b Cos B)² = c²

{Squaring both sides}

=> a²Sin²B +b²Cos²B + 2a.b.Cos B.SinB = c²

-----------(i)

Now,let us assume=

a Cos B - b Sin B = d

On squaring both sides we get =>

(a Cos B - b Sin B)² = d²

=>a²Cos²B + b² Sin²B - 2a.b.Sin B.Cos B= d²

-----------(ii)

Now on adding equation (i) and equation (ii) we get =>

a²Sin²B + b²Cos²B - 2a.b.Sin B.Cos B + a²Cos²B + b² Sin²B - 2a.b.Sin B.Cos B =d² + c²

=> a²Sin²B + b²Cos²B + a²Cos²B + b²Sin²B = d² + c²

=>(Sin²B + Cos²B)a² + (Sin²B + Cos²B)b² = d² + c²

But we know that >

Sin²B + Cos²B = 1

Now on applying this ratio on the expression we get =>

a² + b² = d² + c²

=> a² + b² - c² = d²

=> d = √(a² + b² - c²)

=> a CosB - b SinB = √(a² + b² -c²)

Hence proved.

Remember

1)SinA =$\dfrac{Perpendicular} {Hypotenuse}$

2)CosA=$\dfrac{Base} {Hypotenuse}$

3)TanA=$\dfrac{Perpendicular}{base}$

4)SinA = $\dfrac{1}{Cosec A}$

5)$Cos A =\dfrac{1}{Sec A}$

6)TanA=$\dfrac{1}{Cot A}$

7)$Sin^{2}A+Cos^{2}A=1$

8)$Sec^{2}A-Tan^{2}A=1$

9) $Cosec^{2}A-Cot^{2}A=1$