Question

prove that: tan theta +sec theta -1/tan theta - sec theta +1=1 + sin theta/cos theta

Answer 1

Ok, never mind... I was writing something else.

For reference, you can multiply both the numerator and denominator by \(\cos\theta\).

Answer 2

on lhs or rhs???

Answer 3

\[= \dfrac{\sin\theta +1 - \cos\theta}{\sin\theta - 1 + \cos\theta}\]

Answer 4

On the LHS.

Answer 5

what is cos(tan) and cos(sec)?

Answer 6

\[= \dfrac{\left(\sin \theta + 1 - \cos\theta\right)^2}{(\sin\theta + \cos\theta - 1)(\sin\theta - (\cos\theta-1))}\]Simplify the denominator first separately\[\sin^2\theta - \left(\cos^2\theta + 1 + 2\cos\theta\right)= 1 - \cos^2\theta - \cos^2\theta - 1 - 2\cos\theta = -2\cos\theta(\cos\theta + 1)\]

Answer 7

Remember that since tan = sin/cos, it means that tan * cos = sin.

Answer 8

oh.... yes thank you!

Answer 9

All this question involves is using the identity \(\sin^2\theta + \cos^2\theta = 1\). Just expand all the perfect squares until you get something in the form \(\sin^2 \theta + \cos^2 \theta \)

Answer 10

ya ok! i'll do it! thanks for ur help!!! :)

Answer 11

If you know the identity \( 1 + \tan^2 A = \sec^2 A\), you can use that too.

But all identities are related with each other so just go ahead with any one of them.

Answer 12

ok

Answer 13

ty!

Answer 14

No problem :)

Answer 15

hello! how do u simplify the numerator??

Answer 16

plz help!!!!!!