What I have so far:  11 inches = 0.916667 feet  Let a represent θ  tan a = 0.916667 / 87 = 0.010536402  tan^-1(0.010536402) = 0.6036... rounded to 0.6 degrees.  Is this correct?
121698
In curling, it is often necessary to hit and displace an opponent’s stone to win the end. Olivia would like to hit her opponent’s stone with her own stone. If she releases her stone at the hog line, it needs to travel another 87 feet before reaching her opponent’s stone. Olivia can be off by 11 inches in either direction and still hit her opponent’s stone. Assuming her stone doesn’t curl (change direction), within what angle, $\theta$, to the nearest tenth, must Olivia throw her stone to hit her opponent’s stone?    With this question, do you need to convert the 11 inches to feet, and then solve from there? And since both triangles are the same, do you only solve for one? $11$ inches = $0.916667$ Feet   $\tan (a) = a / b$  $\tan (a) = 0.916667 / 87$  $a = \arctan (0.105364022989)$  $a = 0.6037^\circ$
13838
Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.  Can anyone help me with this? Thank You!
92983
Simplify the expression $$\sin\left(\tan^{-1}(x)\right)$$ Using a triangle with an angle $\theta$, opposite  is x and adjacent is 1 meaning the hypo. is ${\sqrt {x^2+1}}$   Now because the problem has sin after the $\tan^{-1}(x)$ that means that the opp. being x, is put on top of the hypo. giving us the answer, but I feel like I'm missing something because I haven't incorporated the $\tan^{-1}x$
123248
Problem:   Find the value of $$\sum_{r=0}^{44} \tan^2(2r+1)$$  Note: The angles here are in degrees.    I don't know how to solve this question because trigonometric simplifications didn't get me anywhere. I think there was a method to solve this question using complex numbers which I no longer remember. Any hint/help will be appreciated.
31256
Does anyone know of a way to estimate the tangent of an angle in their head? Accuracy is not critically important, but within $5%$ percent would probably be good, 10% may be acceptable.  I can estimate sines and cosines quite well, but I consider division of/by arbitrary values to be too complex for this task. Multiplication of a few values is generally acceptable, and addition and subtraction are fine.  My angles are in degrees, and I prefer not have to mentally convert to radians, though I can if necessary. Also, all angles I'm concerned with are in the range of [0, 90 degrees].  I am also interested in estimating arc tangent under the same conditions, to within about 5-degrees would be good.  Backstory  I'm working on estimating the path of the sun across the sky. I can estimate the declination pretty easily, but now I want to estimate the amount of daylight on any given day and latitude. I've got it down to the arc cosine of the product of two tangents, but resolving the two tangents is now my sticking point. I also want to calculate the altitude of the sun for any time of day, day of the year, and latitude, which I have down to just an arc tangent.
10274
Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
64757
I have $$tan (\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})=\frac{x}{667.6 × 10^6}$$   $$\therefore x = (667.6 × 10^6) \cdot tan(\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})$$   I want to find $x$ with the absolute uncertainty. I don't know how to start, as I am unsure how I would propagate uncertainties with $tan$. Please include your working out when you explain the method of propagating uncertainties in this case.
74654
I'm stuck in this problem. I can not integrate $\int \sec\theta\tan^2\theta \,d\theta$. Can anyone help me.
120874
Prove the following trigonometric identity: $$1 + \tan^2\theta = \sec^2\theta$$  I'm curious to know of the different ways of proving this depending on different characterizations of tangent and secant.
13331
I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer.
116013
I have to prove the following result :  $$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$  I tried converting $\tan\theta$ &amp; $\cot\theta$ into $\cos\theta$ and $\sin\theta$. That led to a huge expression which I wasn't able to simplify.  Please help!!!
65202
Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$    \begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&amp;=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &amp;=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align}  I am stuck here.
120372
If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$   My 1st  approach :   $\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $  $\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution.   My IInd approach :   U\sing Euler substitution :   \since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)  Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$  and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$   By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$  $$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$  Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
57324
I have seen a few solutions but I can't apply them to this particular question. The question is:   Prove that $\cot 142.5° = \sqrt{2} + \sqrt{3} - 2 -  \sqrt{6}$  Help would be really appreciated.
54168
I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do
115154
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$   So, LS= $$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$ $$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$ $$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$ Now, considering the fact that I must have a common denominator to subtract, would this be correct: $$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$ I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
82310
To prove  $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $.  I tried to convert $\tan(4A)$ and $ \tan(8A)$ to $\tan(A$) and tried putting in L.H.S but it becomes a mess. Is there a easy and also intuitive  way to do this?  Thanks