Question

If x= a sin theta , y = b tan theta then prove that a^2/x^2 - b^2/y^2 = 1

Answer 1

since

x=asin@,
then

a/x =1/sin@,

a/x=cosec@,

then

a²/x² =cosec²@,..................eq(1),



since


y=a tan@,

then

a/y =cot@,

hence

a²/y²=cot²@,.......eq(2),


now subtract eq(2) from the eq(1), we have

a²/x² - a²/y²= cosec²@-cot²@,

=1,



since

cosec²@-cot²@=1

Answer 2

Taking L.H.S

a^2/x^2 - b^2/y^2

put the valhlue of X and Y

a^2 / (a sintheta)^2 - b^2 / (b tanthrta)^2

a^2 / a^2 sin^2 theta - b^2 / b^2 tab^2 theta


a^2 se a^ 2 cancel out ho gaya our
b^2 se b^2


1 / sin^2 theta - 1 / tan^2 theta

we know that tan theta= sin / cos


1 / sin^2 theta - 1 / sin^2 theta / cos^2 theta

1 / sin^2 they - cos^2 theta / sin^2 theta


Denomination ( sin^2 theta) is common


then,

1 - cos^2 theta / sin^2 theta


we know that (1 - cos^2 theta=1 )


therefore,. sin^2 theta / sin^2 theta

1

proved