Question

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

Answer 1

Wrong question probably :

Correct question is as :

a sin B + b cos B = c

Upon squaring both sides we get :

⇒ ( a sin B + b cos B )² = c²

Using the formula of expansion of ( a + b )² = a² + b² + 2 ab :

⇒ a² sin²B + b² cos²B + 2 ab sin B cos B = c²

Use the formula of sin²B = 1 - cos²B .

Use the formula of cos²B = 1 - sin²B .

⇒ a²( 1 - cos²B ) + b²( 1 - sin² B ) + 2 ab sin B cos B = c²

⇒ a² - a²cos²B + b² - b² sin²B + 2 ab sin B cos B = c²

Transpose the required values :

⇒ - a²cos²B - b²sin²B + 2 ab sin B cos B = c² - a² - b²

Multiplying both sides by negative signs we get :

⇒ a²cos²B + b²sin²B - 2 ab sin B cos B = a² + b² - c²

Use the formula a² + b² - 2 ab = ( a - b )² :

⇒ ( a cos B - b sin B )² = a² + b² - c²

Taking square roots both sides we get :

⇒ a cos B - b sin B = ± √ ( a² + b² - c² )

Hence the given equation is proved !

Answer 2

Your question is wrong.

Actually it should be >

QUESTION

a Sin B + b Cos B = c

-To Prove

( a CosB - b SinB )= √( a² + b² - c²)

ANSWER

We know that >

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