Question

Perform the indicated expression:
2tan28°tan62°

Answer 1

x= 28
2tan(90-x).tan(x)= 2sin(90-x).sin(x)/cos(90-x).cos(x)

Answer 2

Now sin(90-x)=? cos(90-x)=?

Answer 3

we have this:
\[\Large \tan 90 = \tan \left( {28 + 62} \right) = \frac{{\tan 28 + \tan 62}}{{1 - \tan 28\tan 62}}\]

Answer 4

Ahh, I didn't know you could combine it :)

Answer 5

now, since tan90= infinity, we have to request that:
\[\Large 1 - \tan 28\tan 62 = 0\]

Answer 6

therefore:
\[\Large 2 - 2\tan 28\tan 62 = 0\]

Answer 7

thanks

Answer 8

1/sec^2(36) - cot^(54) =

Answer 9

^ help me with that