Question

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 1

From first equation
\begin{eqnarray*}
\frac{ax}{\cos\theta}+\frac{by}{\sin\theta} & = & a^2-b^2 \\
\frac{ax\sin\theta+by\cos\theta}{\cos\theta\sin\theta} & = & a^2-b^2 \\
ax\sin\theta+by\cos\theta & = & (a^2-b^2)\cos\theta\sin\theta
\end{eqnarray*}
From second equation
\begin{eqnarray*}
\frac{ax\sin\theta}{\cos^2\theta}-\frac{by\cos\theta}{\sin^2\theta} & = & 0 \\
\frac{ax\sin^3\theta-by\cos^3\theta}{\cos^2\theta\sin^2\theta} & = & 0 \\
ax\sin^3\theta-by\cos^3\theta & = & 0 \\
ax\sin^3\theta & = & by\cos^3\theta \\
ax\tan^3\theta & = & by
\end{eqnarray*}
Substituting in first equation:
\begin{eqnarray*}
ax\sin\theta+by\cos\theta & = & (a^2-b^2)\cos\theta\sin\theta \\
ax\sin\theta+(ax\tan^3\theta)\cos\theta & = & (a^2-b^2)\cos\theta\sin\theta \\
ax\sin\theta+ax\tan^2\theta\frac{\sin\theta}{\cos\theta}\cos\theta & = & (a^2-b^2)\cos\theta\sin\theta \\
ax\sin\theta+ax\tan^2\theta\sin\theta & = & (a^2-b^2)\cos\theta\sin\theta \\
ax+ax\tan^2\theta & = & (a^2-b^2)\cos\theta \\
ax(1+\tan^2\theta) & = & (a^2-b^2)\cos\theta \\
ax\sec^2\theta & = & (a^2-b^2)\cos\theta \\
ax & = & (a^2-b^2)\cos^3\theta
\end{eqnarray*}
... to be continued.

Answer 2

continue plzzz

Answer 3

hmmmmm

Answer 4

Hence
\[(ax)^\frac{2}{3} = (a^2-b^2)^\frac{2}{3}\cos^2\theta\]
and
\begin{eqnarray*}
(by)^\frac{2}{3} & = & (ax)^\frac{2}{3}\tan^2\theta \\
& = & ((a^2-b^2)^\frac{2}{3}\cos^2\theta)\tan^2\theta \\
& = & (a^2-b^2)^\frac{2}{3}\cos^2\theta\frac{\sin^2\theta}{\cos^2\theta} \\
(by)^\frac{2}{3} & = & (a^2-b^2)^\frac{2}{3}\sin^2\theta
\end{eqnarray*}
So finally,
\begin{eqnarray*}
(ax)^\frac{2}{3}+(by)^\frac{2}{3} & = & (a^2-b^2)^\frac{2}{3}\cos^2\theta + (a^2-b^2)^\frac{2}{3}\sin^2\theta\\
& = & (a^2-b^2)^\frac{2}{3}(\cos^2\theta+\sin^2\theta) \\
(ax)^\frac{2}{3}+(by)^\frac{2}{3} & = & (a^2-b^2)^\frac{2}{3}
\end{eqnarray*}
as required.

Answer 5

man is this much steps necessary is there an easier way???

Answer 6

Well I tried to make it simpler but I couldn't... :(

Answer 7

wel thx i hav some moe questions plz dont go

Answer 8

k?

Answer 9

Have about 45 more mins here.

Answer 10

wow thx

Answer 11

u ar in which grade

Answer 12

if sin (theta+A)/sin(theta+B)=sqrt(sin 2A/sin2B),prove that tan^2theta=tan( A)*tan (B)

Answer 13

ME??

Answer 14

yes

Answer 15

11th

Answer 16

y u ask??

Answer 17

i am not able to understand it

Answer 18

i am in 9th thats why

Answer 19

I asked why u asked my grade??

Answer 20

u a girl??

Answer 21

yes

Answer 22

got too many fans lol

Answer 23

xactxx do help me

Answer 24

hmmmmm

Answer 25

:)

Answer 26

u have also many fans

Answer 27

i am your fan also

Answer 28

thnk u

Answer 29

welcome

Answer 30

XACTXXX HELPPPPPP

Answer 31

hold on
I first write down the answer on a piece of paper

Answer 32

2.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 33

............

Answer 34

plz answer both and questions over

Answer 35

hmmmmm

Answer 36

ok I got the second one I think
\begin{eqnarray*}
\frac{\sin(\theta+A)}{\sin(\theta+B)} & = &\sqrt{\frac{\sin 2A}{\sin 2B}} \\
\Big(\frac{\sin\theta\cos A+\cos\theta\sin A}{\sin\theta\cos B+\cos\theta\sin B}\Big)^2 & = & \frac{2\sin A\cos A}{2\sin B\cos B} \\
(\sin\theta\cos A+\cos\theta\sin A)^2\sin B\cos B & = & (\sin\theta\cos B+\cos\theta\sin B)^2\sin A\cos A \\
(\sin^2\theta\cos^2 A+2\cos\theta\sin\theta\cos A\sin A+\cos^2\theta\sin^2 A)\sin B\cos B & = & (\sin^2\theta\cos^2 B+2\cos\theta\sin\theta\cos B\sin B+\cos^2\theta\sin^2 B)\sin A\cos A \\
\sin^2\theta\cos^2 A\sin B\cos B+2\cos\theta\sin\theta\cos A\sin A\sin B\cos B+\cos^2\theta\sin^2 A\sin B\cos B & = & \sin^2\theta\cos^2 B\sin A\cos A+2\cos\theta\sin\theta\cos B\sin B\sin A\cos A+\cos^2\theta\sin^2 B\sin A\cos A
\end{eqnarray*}
\begin{eqnarray*}
\sin^2\theta\cos^2 A\sin B\cos B+\cos^2\theta\sin^2 A\sin B\cos B & = & \sin^2\theta\cos^2 B\sin A\cos A+\cos^2\theta\sin^2 B\sin A\cos A \\
\sin^2\theta\cos^2 A\sin B\cos B-\sin^2\theta\cos^2 B\sin A\cos A & = & \cos^2\theta\sin^2 B\sin A\cos A-\cos^2\theta\sin^2 A\sin B\cos B \\
\sin^2\theta\cos A\cos B(\cos A\sin B-\cos B\sin A) & = & \cos^2\theta\sin A\sin B(\cos A\sin B-cos B\sin A) \\
\sin^2\theta\cos A\cos B & = & \cos^2\theta\sin A\sin B \\
\frac{\sin^2\theta}{\cos^2\theta} & = & \frac{\sin A\sin B}{\cos A\cos B} \\
\tan^2\theta & = & \tan A\tan B
\end{eqnarray*}
as required.

Answer 37

can u read???

Answer 38

right click that and click 'show source'. It will help

Answer 39

its not understandable

Answer 40

this show source

Answer 41

\[\tiny\tiny\tiny\tiny\tiny\tiny\small\small\small\small\small{\begin{eqnarray*}
\frac{\sin(\theta+A)}{\sin(\theta+B)} & = &\sqrt{\frac{\sin 2A}{\sin 2B}} \\
\Big(\frac{\sin\theta\cos A+\cos\theta\sin A}{\sin\theta\cos B+\cos\theta\sin B}\Big)^2 & = & \frac{2\sin A\cos A}{2\sin B\cos B} \\
(\sin\theta\cos A+\cos\theta\sin A)^2\sin B\cos B & = & (\sin\theta\cos B+\cos\theta\sin B)^2\sin A\cos A \\
(\sin^2\theta\cos^2 A+2\cos\theta\sin\theta\cos A\sin A+\cos^2\theta\sin^2 A)\sin B\cos B & = & (\sin^2\theta\cos^2 B+2\cos\theta\sin\theta\cos B\sin B+\cos^2\theta\sin^2 B)\sin A\cos A \\
\sin^2\theta\cos^2 A\sin B\cos B+2\cos\theta\sin\theta\cos A\sin A\sin B\cos B+\cos^2\theta\sin^2 A\sin B\cos B & = & \sin^2\theta\cos^2 B\sin A\cos A+2\cos\theta\sin\theta\cos B\sin B\sin A\cos A+\cos^2\theta\sin^2 B\sin A\cos A
\end{eqnarray*}}\]
I've tried my best

Answer 42

AravindG, right click then choose Zoom Trigger, Click, then click on the mathematical text. You could scroll through the solution afterwards.

Answer 43

still right part is missing