Answer:
[tex](c)\ \tan B = \sin A[/tex]
Step-by-step explanation:
Given
[tex]\angle A + \angle B = 90[/tex] --- Complement angles
See attachment for complete question
Required
Which is not always true
To do this, we simply test each option
[tex](a)\ \sin A = \cos B[/tex]
The above is always true, if A and B are complements.
Examples are:
[tex]\sin(40) = \cos(50)[/tex]
[tex]\sin(90) = \cos(0)[/tex]
etc
[tex](b)\ \sec A = \csc B[/tex]
The above is always true, if A and B are complements.
The expression can be further simplified as:
[tex]\frac{1}{\cos A} = \frac{1}{\sin B}[/tex]
Cross Multiply
[tex]\sin B = \cos A[/tex]
This is literally the same as (a)
[tex](c)\ \tan B = \sin A[/tex]
The above is not always true, if A and B are complements.
The expression can be further simplified as:
[tex]\frac{\sin B}{\cos B} = \sin A[/tex]
Cross multiply
[tex]\sin B = \sin A * \cos B[/tex]
If A and B are complements. then
[tex]\sin A = \cos B[/tex]
So, we have:
[tex]\sin B = \sin A * \sin A[/tex]
[tex]\sin B = \sin^2 A[/tex]
The above expression is not true, for values of A and B
[tex](d) \cot B = \tan A[/tex]
The above is always true, if A and B are complements.
An example is:
[tex]\cot (55) = \tan (25) = 0.7002[/tex]
etc.
