Establish the following identity
Sin2x/1+cos2x=tanx
To establish the identity π ππ^2π₯/(1+πππ ^2π₯) = π‘πππ₯, we will use the trigonometric identity π‘ππ^2π₯ + 1 = π ππ^2π₯.
Start with the left-hand side (LHS) of the equation:
π ππ^2π₯/(1+πππ ^2π₯)
Using the identity π‘ππ^2π₯ + 1 = π ππ^2π₯, we can rewrite the denominator π ππ^2π₯/(1+πππ ^2π₯) as:
π ππ^2π₯/(π‘ππ^2π₯) (By substituting 1+cos^2π₯ with π‘ππ^2π₯)
Next, using the identity π ππ^2π₯ = 1βπππ ^2π₯, we can further simplify the numerator:
(1βπππ ^2π₯)/(π‘ππ^2π₯) (By substituting π ππ^2π₯ with 1βπππ ^2π₯)
Since π‘ππ^2π₯ is the reciprocal of πππ ^2π₯, we can further simplify:
(1βπππ ^2π₯)/(1/πππ ^2π₯) (By substituting π‘ππ^2π₯ with 1/πππ ^2π₯)
To simplify the expression above, we multiply the numerator and denominator by πππ ^2π₯:
(1βπππ ^2π₯)/(1/πππ ^2π₯) * πππ ^2π₯/πππ ^2π₯
= (1βπππ ^2π₯)πππ ^2π₯
Using the identity (1βπππ ^2π₯)πππ ^2π₯ = π ππ^2π₯, we have:
π ππ^2π₯
Thus, the LHS simplifies to π ππ^2π₯, which is equal to the right-hand side (RHS) of the equation π‘πππ₯.
Hence, we have established the identity π ππ^2π₯/(1+πππ ^2π₯) = π‘πππ₯.