### Video Transcript

True or False: In the given figure, π
sub one over the cos of 90 degrees minus π is equal to π
sub two over sin π, which is equal to π over sine of 90 degrees.

The figure shows two perpendicular vector forces π
sub one and π
sub two, together with their resultant π which acts at an angle of π degrees to the horizontal. Since the line segments ππ΅ and π΄πΆ are parallel and equal in length and as vectors describe direction and magnitude but not location, we can let the line segment π΄πΆ be equal to the vector π
sub two. We can now consider triangle ππ΄πΆ in order to find the link between vector π
sub one, π
sub two, and their resultant.

Since π
sub one and π
sub two are perpendicular, we know that angle ππ΄πΆ is a right angle and is therefore equal to 90 degrees. We know that angles in a triangle sum to 180 degrees. This means that the measure of angle ππΆπ΄ is equal to 180 degrees minus 90 degrees plus π, which is equal to 90 degrees minus π.

Next, we can use the sine rule or law of sines. This states that π over sin π΄ is equal to π over sin π΅, which is equal to π over sin πΆ, where lowercase π, π, and π are the lengths of the three sides of our triangle and capital π΄, π΅, and πΆ are the angles opposite the corresponding sides. In our triangle, vector π
sub one is opposite the angle 90 degrees minus π, π
sub two is opposite angle π, and the resultant force π is opposite the angle of 90 degrees. Substituting in these values gives us π
sub one over sin of 90 degrees minus π is equal to π
sub two over sin π, which is equal to π over sin of 90 degrees.

The second and third parts of our equation are correct. However, the denominator of the first part of the equation in the question was the cos of 90 degrees minus π, whereas in the figure the correct expression is the sin of 90 degrees minus π. From our knowledge of supplementary angles and the trigonometric identities, we know that the cos of 90 degrees minus π is equal to the sin of π, and this will only be the same as sin of 90 degrees minus π when π is 45 degrees.

If π was equal to 45 degrees, then ππ΄πΆπ΅ would be a square, and the magnitude of π
sub one would be equal to the magnitude of π
sub two. As we are not told that this is the case in this question, we can conclude that the statement is false.