Question

Q. if (3sinx + 5cosx) = 5 ,then the value of (5sinx - 3cosx ) is (a)4 ,(b)3 ,(c)2 d(1)

Answer 1

Answer:

Option (b) is correct

Step-by-step explanation:

Given 3sinx+5cosx=5 ----(1)

On Squaring both sides of the equation (1) , we get

(3sinx+5cosx)²=5²

=> (3sinx)²+(5cosx)²+2(3sinx)(5cosx)=25

$By\: algebraic\: identity:\:\\</p><p>\boxed{(a+b)^{2}=a^{2}+b^{2}+2ab}$

=> 9sin²x+25cos²x+2(3sinx)(5cosx)=25

=> 9(1-cos²x)+25(1-sin²x)+2(3sinx)(5cosx)=25

$By\: Trigonometric\: identities:\:\\</p><p>\boxed{i)sin^{2}A=1-cos^{2}A\\ii)cos^{2}A=1-sin^{2}A}$

=> 9-9cos²x+25-25sin²x+2(3sinx)(5cosx)=25

=> -9cos²x-25sin²x+2(3sinx)(5cosx)=25-9-25

=> -9cos²x-25sin²x+2(3sinx)(5cosx)=-9

On multiplying both sides by (-1) , we get

=> 9cos²x+25sin²x-2(3sinx)(5cosx)=9

=> (3cosx)²+(5sinx)²-2(3cosx)(5sinx)=3²

=> (3cosx-5sinx)²=3²

$By\: algebraic\: identity:\:\\</p><p>\boxed{(a-b)^{2}=a^{2}+b^{2}-2ab}$

=> 3cosx-5sinx =±3

According to the options given,

Option (b) is correct

•••♪

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given 3sinx+5cosx=5 ----(1)

On Squaring both sides of the equation (1) , we get

(3sinx+5cosx)²=5²

=> (3sinx)²+(5cosx)²+2(3sinx)(5cosx)=25

\begin{lgathered}By\: algebraic\: identity:\:\\ \boxed{(a+b)^{2}=a^{2}+b^{2}+2ab}\end{lgathered}

Byalgebraicidentity:

(a+b)

2

=a

2

+b

2

+2ab

=> 9sin²x+25cos²x+2(3sinx)(5cosx)=25

=> 9(1-cos²x)+25(1-sin²x)+2(3sinx)(5cosx)=25

=> 9-9cos²x+25-25sin²x+2(3sinx)(5cosx)=25

=> -9cos²x-25sin²x+2(3sinx)(5cosx)=25-9-25

=> -9cos²x-25sin²x+2(3sinx)(5cosx)=-9

On multiplying both sides by (-1) , we get

=> 9cos²x+25sin²x-2(3sinx)(5cosx)=9

=> (3cosx)²+(5sinx)²-2(3cosx)(5sinx)=3²

=> (3cosx-5sinx)²=3²

\begin{lgathered}By\: algebraic\: identity:\:\\ \boxed{(a-b)^{2}=a^{2}+b^{2}-2ab}\end{lgathered}

Byalgebraicidentity:

(a−b)

2

=a

2

+b

2

−2ab

=> 3cosx-5sinx =±3