4tanA%1-tan^4A=tan2A+sin2A

See prove of the identity below

Step-by-step explanation:

Recall that the following difference can be factored out as:

Now we look at the right side of the equal sign, and recall the identity that relates the sine of a double angle 2A with an expression in terms of the tan(A):

and similarly, the tangent of the double angle "2A" can be written as:

We can then combine the two expressions on the right using  the common denominator: :

And we have proved the identity since the right hand side becomes exactly equal to the left hand side of the equal sign.

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake