trigonoo

In the RHS substituting the values of m and n from what is already given in the question, we get ::

\frac{\frac{sin^{2}A}{sin^{2}B}-1}{\frac{tan^{2}A}{tan^{2}B}-1}=\left( \frac{sin^{2}A-sin^{2}B}{tan^{2}A-tan^{2}B}\right)\left(\frac{tan^{2}B}{sin^{2}B} \right)

=\left( \frac{sin^{2}A-sin^{2}B}{\left(\frac{sin^{2}A}{cos^{2}A} \right)-\left(\frac{sin^{2}B}{cos^{2}B} \right)\right)}\left(\frac{1}{cos^{2}B} \right)

=\left(\frac{sin^{2}A-sin^{2}B}{\frac{sin^{2}A}{cos^{2}A}.cos^{2}B - sin^{2}B} \right)=\left(\frac{sin^{2}A-sin^{2}B}{sin^{2}A.cos^{2}B - sin^{2}B.cos^{2}A} \right)cos^{2}A

=\left(\frac{sin^{2}A-sin^{2}B}{\frac{sin^{2}A}{cos^{2}A}.cos^{2}B - sin^{2}B} \right)=\left(\frac{sin^{2}A-sin^{2}B}{sin^{2}A.(1-sin^{2}B) - sin^{2}B.(1-sin^{2}A)} \right)cos^{2}A = cos^{2}A

( in the last term expand the denominator, terms get cancel out to give cos ² A

just divide 2nd by 1snd then square it on both the sides..

work on the 2nd on to find
cos2b nd then subs it in
that eqn to get cos2a.....

this reduces calculation...≠just divide 2nd by 1snd then square it on both the sides..

work on the 2nd on to find
cos2b nd then subs it in
that eqn to get cos2a.....

this reduces calculation...

divide second by first......
square it.....
let thiis eqn be *1
then find the value of cos^2 B from
second eqn...
substitute it in eqn *1 to get
cos^2 A

....... sorry fr the above three posts...
i m nt getting the option for removing
it in my cellphone....