If alpha and beta are solutions of acostheta bsintheta

Asked by admin @ 23/08/2021 in Math viewed by 321 People

if alpha and beta are two distinct roots satisfying the equation acostheta +bsintheta =c prove that sin (alpha+beta) =2ab/asquare+bsquare

Answered by admin @ 23/08/2021

Answer:

Step-by-step explanation:

Given,

$a\cos\theta+b\sin\theta=c\\\;\\\text{Taking}\;\cos\theta\;\text{common}}\\\;\\a+b\frac{\sin\theta}{\cos\theta}=\frac{c}{\cos\theta}\\\;\\a+b\tan\theta=c\sec\theta\\\;\\\text{Making square of both sides,}\\\;\\(a+b\tan\theta)^2=(c\sec\theta)^2\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2\sec^2\theta\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2(1+\tan^2\theta)\;\;\;\;\;\;\;\;(\sec^2\theta=1+\tan^2theta)\\\;\\a^2+b^2\tan^2\theta+2ab\tan\theta=c^2+c^2\tan^2\theta\\\;\\(b^2-c^2)\tan^2\theta+2ab\tan\theta+a^2-c^2=0$

This is in a form of a quadratic equation, whoes roots are α and β. If we compare this equation with standard from, We find that x is compared to tan∅. hence, We can say that tanα and tanβ are solutions of this quadratic equation.

Now,

$\text{Sum of roots}=-\frac{2ab}{b^2-c^2}\\\;\\\tan\alpha+\tan\beta=-\frac{2ab}{b^2-c^2}\;\;\;\;\;\;\..................i)\\\;\\\;\\\text{Product of roots}=\frac{a^2-c^2}{b^2-c^2}\\\;\\\tan\alpha.\tan\beta=\frac{a^2-c^2}{b^2-c^2}\;\;\;\;\;..............ii)$

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha.\tan\beta}\\\;\\\tan(\alpha+\beta)=\frac{\frac{-2ab}{b^2-c^2}}{1-\frac{a^2-c^2}{b^2-c^2}}\\\;\\\tan(\alpha+\beta)=\frac{-2ab}{b^2-c^2-a^2+c^2}\\\;\\\tan(\alpha+\beta)=\frac{2ab}{b^2-a^2}$

$1+\tan^2(\alpha+\beta)=1+(\frac{2ab}{b^2-a^2})^2\\\;\\1+\tan^2(\alpha+\beta)=\frac{b^4+a^4-2a^2b^2+4a^2b^2}{(b^2-a^2)^2}\\\;\\1+\tan^2(\alpha+\beta)=\frac{a^4+b^4+2a^2b^2}{(b^2-a^2)^2}\\\;\\1+\tan^2(\alpha+\beta)=\frac{(a^2)^2+(b^2)^2+2a^2b^2}{(b^2-a^2)^2}\\\;\\1+\tan^2(\alpha+\beta)=\frac{(a^2+b^2)^2}{(b^2-a^2)^2)}\\\;\\\sqrt{1+\tan^2(\alpha+\beta)}=\frac{a^2+b^2}{b^2-a^2}$

Now,We know that,

$\sin(\alpha+\beta)=\frac{\tan(\alpha+\beta)}{\sqrt{1+\tan^2(\alpha+\beta)}}\\\;\\\sin(\alpha+\beta)=\frac{\frac{2ab}{b^2-a^2}}{\frac{a^2+b^2}{b^2-a^2}}\\\;\\\sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}\\\;\\\textbf{Hence Proved}$

If alpha and beta are solutions of acostheta bsintheta

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