Question

.If alpha and beta be 2 distinct roots satisfying equation a cos theta+bsin theta=c,Show that cos (alpha+beta)=(a^2-b^2)/(a^2+b^2)

Answer 1

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Answer 2

\[a\cos(\theta)+b\sin(\theta)=c\]
prove that
\[\cos(\alpha+\beta)=\frac{a^2-b^2}{a^2+b^2}\]

Answer 3

ya

Answer 4

OKay...I am posting the solution ..I got it

Answer 5

\[\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\]

Answer 6

\[a\cos(\alpha)+b\sin(\alpha) = a\cos(\beta)+b\sin(\beta)\]

Answer 7

\[cos(\alpha+\beta)= cos\alpha*cos\beta - sin\alpha*sin\beta\]
\[acos\theta + bsin\theta=c\]
\[acos\theta = c - bsin\theta\]
\[\text{Square both sides}\]
\[a^2 (1 - sin^\theta) = c^2 + b^2sin^2\theta -2*c*b*sin\theta\]
\[(b^2 + a^2 )sin^2\theta -2 *c*b*sin\theta + c^2 -a^2=0\]
\[\text{Product of Roots }= \frac{c}{a}\]
\[sin\alpha*sin\beta=\frac{c^2-a^2}{a^2+b^2}\]
\[\text{Now use similar method to have an equation in terms of } cos^2\theta\]
\[acos\theta + bsin\theta=c\]
\[acos\theta-c = -bsin\theta\]
\[a^2cos^2\theta + c^2 - 2*a*ccos\theta = b^2 (1-cos^2\theta)\]
\[(a^2 + b^2)cos^2\theta - 2*a*cos\theta*c +c^2 -b^2=0\]
\[\text{Product of Roots }= \frac{c}{a}\]
\[cos\alpha*cos\beta = \frac{c^2-b^2}{a^2+b^2}\]
\[\text {Now substitute the values in First Equation }\]

\[cos(\alpha+\beta)= cos\alpha*cos\beta - sin\alpha*sin\beta\]
\[\implies \frac{c^2 -b^2 + a^2 -c^2}{a^2+b^2}\]\[\implies \frac{a^2-b^2}{a^2+b^2}\]
\[\text{Your Question is Done : )}\]

Answer 8

wow...Looks so Beautiful ..\(\LaTeX\)

Answer 9

Aravind !!!!

Answer 10

thnk youuuuuuuuuuuuuuuuuuuuu

Answer 11

ax/cos theta)+(by/sin theta)=a^2-b^2 and (axsin theta/cos^2 theta)-(bycos theta/sin^2 theta)=0
show that (ax)^2/3 +(by)^2/3=(a^2-b^2)^2/3

Answer 12

can u help this one ???i got answer by taking long steps ...bt i feel there is an easier way out can u help??