Question

3sina+4cosa=5 find 4sina-3cosa=

Answer 1

l think it may help you.

Answer 2

$4\sin A-3\cos A=0$

Step-by-step explanation:

We have,

$3\sin A+4\cos A=5$               ......(1)

Let $4\sin A-3\cos A=x$           ......(2)

To find, the vaue of $4\sin A-3\cos A=?$

Squaring and adding (1) and (2), we get

$(3\sin A+4\cos A)^{2}+(4\sin A-3\cos A)^{2}=5^{2}+x^{2}$

⇒ $(3\sin A)^2+(4\cos A)^{2}+2(3\sin A)(4\cos A)+(4\sin A)^2+(3\cos A)^{2}-2(4\sin A)(3\cos A)=25+x^{2}$

⇒ $(9\sin^2 A+16\cos^{2} A+24\sin A\cos A)+(16\sin^2 A+9\cos^{2}-24\sin A\cos A)=25+x^{2}$

⇒ $9\sin^2 A+16\cos^{2} A+16\sin^2 A+9\cos^{2}=25+x^{2}$

⇒ $9(\sin^2 A+\cos^{2} A)+16(\sin^2 A+\cos^{2} A)=25+x^{2}$

⇒ $9(1)+16(1)=25+x^{2}$

⇒ $25=25+x^{2}$

⇒ $x^{2}=25-25=0$

⇒ x = 0

Hence, $4\sin A-3\cos A=0$