Question

please help:

Answer 1

@phi @TheSmartOne @sweetburger @pooja195

Answer 2

I would change tangent into sin/cos
and do the same for cotangent. Do you know how ?

Answer 3

i got tan^2 theta

Answer 4

@phi

Answer 5

yes, that looks ok
\[ \frac{ \frac{\sin x }{\cos x}}{ \frac{\cos x}{\sin x }} \\
\]
multiply top and bottom by sinx / cosx
you get
\[ \frac{ \frac{\sin^2 x }{\cos^2 x}}{1}\]
which is the same as
\[ \frac{\sin^2 x }{\cos^2 x} = \left(\frac{\sin x }{\cos x} \right)^2 = \tan^2 x \]

Answer 6

thanks! @phi can I have help with one more please?

Answer 7

did you make a new post ?

Answer 8

It would have been easier if you used the fact that cot = 1/tan

\(\Large \frac{tan\theta}{cot\theta} = tan\theta \div cot \theta = \)

\(\Large tan\theta\div \frac{1}{tan \theta} = tan \theta \times tan \theta = \boxed{\bf{tan^2\theta}}\)