Question

hi i hav some trigonometry questions can anyone help????

Answer 1

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

Answer 2

it works like this... you post a problem... then someone helps

Answer 3

pls hurry-up arvind

Answer 4

k

Answer 5

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 6

help guys

Answer 7

can u use the equation from

Answer 8

k

Answer 9

its here

Answer 10

http://openstudy.com/groups/mathematics/updates/4e630db70b8b1f45b4ab53d0

Answer 11

it works like this... you post a problem... then someone helps

Answer 12

it's m l khanna problem
\[\sin^2\Theta*\sin \Theta/\cos^2\Theta*\cos \Theta=by/ax\]
\[\sin^3\Theta/\cos^3\Theta=by/ax\]
\[\tan^3\Theta=by/ax\]
\[\tan \Theta=(by/ax)^1/3\]
\[\sin \Theta=(by)^{1/3}/\left[ (ax)^{2/3+(by)^{2/3}} \right]^{1/2}\]
\[\cos \Theta=(ax)^{1/3}/\left[ (ax)^{2/3+(by)^{2/3}} \right]^{1/2}\]
putting the value sin theta and cos theta
\[\left[ (ax)^{2/3+(by)^{2/3}} \right]^{3/2\left[ (ax)^{2/3} \right]}\]
(ax)^2/3 +(by)^2/3=(a^2-b^2)^2/3 (proved)

M.L Khanna & J.SHARMA pg no 320