Help with trig verification/ID's!!!

[lsd]

sec^2(Θ)+sec(Θ)tan(Θ)-1


Now, I'm given these answers that it could be any of...


sec(Θ)
-sec(Θ)
cost(Θ)
-cos(Θ)
csc(Θ)
-csc(Θ)
sin(Θ)
-sin(Θ)
1
(1-sin(Θ))/(sin(Θ))
(sin(Θ)+1)/(sin(Θ))
sin(Θ)/(1+sin(Θ))
(sin(Θ)-1)/(sin(Θ))
sin(Θ)/(1-sin(Θ))
0
-1



Anyone? I get to sin(sinΘ+1)/(1-sin^2(Θ))

Am I close?

 

[soroban]

Hello, lsd!

\(\displaystyle \text{Simplify: }\:\sec^2\theta +\sec\theta\tan\theta -1\)


\(\displaystyle \text{Answer choices:}\)

\(\displaystyle (1)\;\sin\theta \quad (2)\;\text{-}\sin\theta \quad (3)\;\cos\theta \quad (4)\;\text{-}\cos\theta \quad (5)\;\sec\theta \quad (6)\;\text{-}\sec\theta\) . . \(\displaystyle (7)\; \csc\theta \quad (8)\;\text{-}\csc\theta \quad (9)\;1 \)

\(\displaystyle (10)\;\dfrac{1-\sin\theta}{\sin\theta} \quad (11)\;\dfrac{\sin\theta +1}{\sin\theta} \quad (12)\;\dfrac{\sin\theta}{1+\sin\theta} \quad (13)\;\dfrac{\sin\theta -1}{\sin\theta}\) . . \(\displaystyle (14)\;\dfrac{\sin\theta}{1-\sin\theta} \quad (15)\;0 \quad (16)\;\text{-}1 \)

\(\displaystyle \sec^2\!\theta + \sec\theta\tan\theta - 1 \;=\;(\sec^2\!\theta - 1) + \sec\theta\tan\theta \;=\;\tan^2\!\theta + \sec\theta\tan\theta\)

. . . . . . . . . . . . . . . . . \(\displaystyle \displaystyle=\;\frac{\sin^2\theta}{\cos^2\theta} + \frac{1}{\cos\theta}\!\cdot\!\frac{\sin\theta}{\cos\theta} \;=\;\frac{\sin^2\!\theta + \sin\theta}{\cos^2\!\theta} \;=\;\frac{\sin\theta(1 + \sin\theta)}{\cos^2\!\theta}\)


\(\displaystyle \text{Multiply by }\dfrac{1-\sin\theta}{1-\sin\theta}:\;\;\dfrac{\sin\theta(1 + \sin\theta)}{\cos^2\!\theta} \cdot\dfrac{1-\sin\theta}{1-\sin\theta}\) . \(\displaystyle =\;\dfrac{\sin\theta(1 - \sin^2\!\theta)}{\cos^2\!\theta(1-\sin\theta)} \)

. . . . . . . . . . . . . . . . \(\displaystyle =\;\dfrac{\sin\theta\cos^2\!\theta}{\cos^2\!\theta(1-\sin\theta)} \;=\; \dfrac{\sin\theta}{1 - \sin\theta}\;\;\text{ . . . answer (14)}\)

 

[lsd]

Another problem

Thank you a lot!

I have another problem, if you could:

- [(cot²(Θ))/(sec²(Θ)-tan²(Θ)-csc(Θ)]

I get as far as:

1-csc²(Θ)/(-1+csc(Θ))


Answers same choices as the first one.

 

[lsd]

Bump. Anyone?

 

[soroban]

Hello again, lsd!

\(\displaystyle \text{Simplify: }\:- \dfrac{\cot^2\theta}{\sec^2\theta -\tan^2\theta -\csc\theta}\)

\(\displaystyle \text{I get as far as: }\:\dfrac{1-\csc^2\theta}{-1+\csc\theta}\) . . . . no

\(\displaystyle \text{You have: }\:-\dfrac{\cot^2\theta}{1 - \csc\theta} \;=\;\dfrac{\cot^2\theta}{\csc\theta -1} \;=\;\dfrac{\csc^2\theta - 1}{\csc\theta - 1}\) .\(\displaystyle =\;\dfrac{(\csc\theta -1)(\csc\theta+1)}{\csc\theta-1}\)

. . . . . . . . \(\displaystyle =\;\csc\theta + 1 \;=\;\dfrac{1}{\sin\theta} + 1 \;=\;\dfrac{1 + \sin\theta}{\sin\theta}\;\text{ . . . Answer (11)}\)

 

[Deleted member 4993]

Bump. Anyone?

Start a new thread with a new problem .... that way you'll get faster response