Question

Verify the identity.

tan (x + (pi/2)) = -cot x

Answer 1

\[\tan \left( x +\frac{ \Pi }{ 2 } \right)=-\cot x\]

Answer 2

break tan into sin and cos

Answer 3

how do I do that

Answer 4

tanx = (sin x)/ (cos x)

Answer 5

hold on I will give you the solution

Answer 6

thanks, but will you show the work so I can try the next one myself?

Answer 7

If you use the definition of \(\tan x = \sin x/\cos x\) you can rewrite the problem as
\[\tan(x+\frac{\pi}{2}) = \frac{\sin(x+\frac{\pi}{2})}{\cos(x+\frac{\pi}{2})}\]

Next, use the formulas for the sin and cos of a sum of angles:
\[\sin(A+B) = \sin A\cos B + \cos A\sin B\]\[\cos(A+B) = \cos A\cos B - \sin A\sin B\] to simplify that expression. Finally, remember that \[\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}\]

Answer 8

or you could use tan (A+B) formula
\(\tan (A+B)= \dfrac{\tan A +\tan B}{1-\tan A
tan B}\)
plug in A =x, B= pi/2

Answer 9

or use the cofunction identities of
$$
sin\pmatrix{x+\frac{\pi}{2}} = cos(x)\\
cos\pmatrix{x+\frac{\pi}{2}} = -sin(x)
$$

Answer 10

you woulf find the solution in the attachment

Answer 11

@hartnn the problem with that formula is that tan (pi/2) is infinity

Answer 12

yes, yes! thats why my next step was to divide the numerator and denominator by 'tan B'
but that would be \(\large if\) the asker would have tried it, got stuck and asked...

Answer 13

ok you make sense I just did it in my mind nice one

Answer 14

@hartnn