if sin2x=3sin2y,
prove that:
2tan(x-y)=tan(x+y)
( here, in sin2x, 2x is an angle., like there's a formula:sin2x=2sinxcosx and sin2y=2sinycosy; ....)
if sin2x=3sin2y
then sin2x/sin2y=3/1
sin2x+sin2y/sin2x-sin2y=3+1/3-1 (using conponendo and dividend rule)
now using formula of sin2x+sin2y and sin2x-sin2y
2sin (x+y)×cos (x-y)/2cos (x+y)×sin (x-y)=4/2
solving this we get the result
give the proper solution for it. not by substituting values of x and y
I made a mistake in my example. Choose the angles x = 15 degrees and y = 4.79703 degrees. Then the condition
sin 2x = 3 sin 2y is satisfied.
2tan(x-y) = 0.35996
tan (x+y) = 0.35996
Your identity is almost certainly correct and I must apologize for my error in the previous answer.
I tried to prove the second equation for any x and y that satisfy your first equation, but was not able to.
If sin 2x = 3 sin 2y then show that 2 tan ( x _ y) = ( x + y)
If Sin2x=3sin2y then show that 2tan(x-y)=tan(x+y).
sin2x=3sin2y then prove 2tan(x-y)=tan(x+y)
It cannot be proven because it is not true. You can prove that by choosing x and y that satisfies sin2x=3sin2y. For example, take x = 15 degrees and y = 9.5941 degrees.
For those choices, sin 2x = 3sin 2y = 0.500
For those choices of x and y
2 tan(x-y) = 0.1893 and tan (x+y) = 0.4577
By being able to find ANY x,y combination that satisfies sin2x=3sin2y, and for which your second equation is not true, then your contention is disproved.
Either someone is playing a practical joke on us, or on you, or you have copied the problem incorrectly