prove that [(sin 2 t / sin t )] - [( cos 2t ) / cos t ] = sec t
and sin (2t - t ) = sin t
On the first, start with the double angle formulas.
On the second, the proof is trivial.
2t-t is t.
sin(2t-t)=sint
In fact I have solved this like that
follows:
=(sin 2t*cost-cos2t*sint )/cost*sint
= sin (2t-t)/sint*cost
= sint/sint*cost
=1/cost
=sect QED
Where 't' stands for 'theta' .
Was I correct ?
To prove the first equation, let's simplify the left-hand side:
[(sin 2t / sin t) - (cos 2t / cos t)]
First, let's rationalize the denominators:
[(sin 2t * cos t - cos 2t * sin t) / (sin t * cos t)]
Using the angle subtraction formula for sine:
[(sin (2t - t)) / (sin t * cos t)]
Since sin (2t - t) is equal to sin t:
[(sin t) / (sin t * cos t)]
Now, cancel out the common factor of sin t:
[1 / cos t]
Which is equal to sec t.
Therefore, [(sin 2t / sin t) - (cos 2t / cos t)] is equal to sec t.
Now let's prove the second equation:
sin (2t - t) = sin t
Using the angle subtraction formula for sine:
sin 2t - t = sin t
Since sin 2t - t is equal to sin t:
sin t = sin t
Therefore, sin (2t - t) is equal to sin t.
To prove the given expressions, we will use trigonometric identities.
Proof for the first expression:
We start with the left side of the equation:
[(sin 2t / sin t) - (cos 2t / cos t)]
To simplify this expression, we need to find a common denominator. The common denominator for sin t and cos t is sin t * cos t.
Rewriting the expression with the common denominator:
[(sin 2t * cos t - cos 2t * sin t) / (sin t * cos t)]
Now, let's apply the trigonometric identity: sin(A - B) = sin A * cos B - cos A * sin B.
Replacing A with 2t and B with t in the numerator:
[sin (2t - t) / (sin t * cos t)]
Using the given identity sin (2t - t) = sin t:
[sin t / (sin t * cos t)]
Now we can cancel out the sin t in the numerator and denominator:
[1 / cos t]
Since 1 / cos t is equal to sec t, we have proved that:
[(sin 2t / sin t) - (cos 2t / cos t)] = sec t
Proof for the second expression:
We need to prove that sin(2t - t) = sin t.
Using the trigonometric identity sin(A - B) = sin A * cos B - cos A * sin B, we can rewrite the expression:
sin(2t - t) = sin 2t * cos t - cos 2t * sin t
Now, let's simplify this expression:
sin(2t - t) = sin t * cos t + cos t * sin t
Combining like terms:
sin(2t - t) = 2sin t * cos t
Using the double angle identity for sine, sin(2t) = 2sin t * cos t:
sin(2t - t) = sin(2t)
Since sin(2t - t) = sin(2t), we have proved that:
sin(2t - t) = sin t