Question

A cos theta + b Sin theta is equals to m and a sin theta minus B cos theta is equals to N prove that a square + b square is equals to m square + n square​

Answer 1

Answer:

Hence proved $a^{2}+b^{2}=m^{2}+n^{2}$.

Step-by-step explanation:

The given equations are:

$aCos\theta + bSin\theta=m\\aSin\theta - bCos\theta=n$

Square both the equations and add:

$(aCos\theta + bSin\theta)^{2}=(m)^{2}\\(aSin\theta - bCos\theta)^{2}=(n)^{2}$

⇒

$a^{2}Cos^{2}\theta+b^{2}Sin^{2}\theta+2abCos\theta Sin\theta=m^{2}\\a^{2}Sin^{2}\theta+b^{2}Cos^{2}\theta-2abCos\theta Sin\theta=m^{2}$

⇒

$a^{2}(Cos^{2}\theta+Sin^{2}\theta)+b^{2}(Sin^{2}\theta+Cos^{2}\theta)=m^{2}+n^{2}$

According to trigonometric identities: $Sin^{2}\theta+Cos^{2}\theta=1$

⇒

$a^{2}+b^{2}=m^{2}+n^{2}$

Hence proved.

Answer 2

Step-by-step explanation:

hope this helps!!

u r given with two identities take squaring of both then u will get another two equations then normally add them and use first identify of trigonometry that is sin square theta plus cos square theta is equals to 1.