How do I myself start off this question:     Show that: $$\frac{\tan x+\sec x-1}{\tan x-\sec x+1}=\frac{1+\sin  x}{\cos x}$$   I have tried to express the LHS in terms of $\sin x$ and $\cos x$,  simplified the resulting expression, squared numerator and denominator  and simplified again, by this time I was lost in the forest…
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Prove: $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$$  My attempt:  LHS= $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$$ $$=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}$$ $$=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$ $$\text{Using componendo and dividendo}$$ $$\frac{\sin A+1-\cos A+\sin A-1+\cos A}{\sin A+1-\cos A-\sin A+1-\cos A}$$ $$\frac{2\sin A}{2-2\cos A}=\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}$$ $$\frac{\sin A(1+\cos A)}{\sin^2 A}=\frac{1+\cos A}{\sin A}$$ $$\text{Which is not equal to right hand side!}$$  Have I done something wrong in the componendo dividendo step? I dont know how to use componendo dividendo rule . I saw it being used like this in some question and hence applied it here the same way. Maybe I am wrong in application of that rule. Please tell me the right way to use it. Thank you.
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I'm doing some math exercises but I got stuck on this problem.  In the book of Bogoslavov Vene it says to prove that:  $$\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.$$  It is easy if we do it like this: $\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x$. But how to prove it just by going from the left part or from the right  Can anybody help me?  Thank you!
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Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?  Can someone point me to a proof, or explain if it's a simple answer?  What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series.  A lot of the information I find when looking for that seems to point back to arctan.
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Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$   There's a solution  that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.   My proposed solution   $$\begin{align} &amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&amp;\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&amp;\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align} $$  And this implies  $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
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I am stuck the following problem :  Solve the following problem: $(1-\tan^2x)\sec^2x+2^{\tan^2x}=0$  Taking $p=\tan^2x,$ we get $(1-p^2)+2^p=0$. Now,I do not know how to go from here. Simple observation says $x=2n\pi \pm \frac{\pi}{3}$,satisfies the equation and so it will be the solution but I do not know how to get there. Can someone help?
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If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$    $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$ $\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})^3$ Applying componendo and dividendo rule on both sides, $\tan\frac{y}{2}=\frac{\tan\frac{x}{2}(3+\tan^2\frac{x}{2})}{(1+3\tan^2\frac{x}{2})}$ I am stuck here.
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I'm having problem with showing that: $$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$  I would need some help in the right direction
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I have got some trouble with proving that for $x\neq 0$: $$ \frac{\arctan x}{x }&lt; 1 $$ I tried doing something like $x = \tan t$ and playing with this with no success.
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We have to evaluate the following integral:  $$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$  I tried this:  I multiplied both the numerator and denominator by $\sec x$ And substituted $\tan x = t$.  But after that I got stuck.  The book where this is taken from gives the following as the answer: $$\ln(1+t)-\frac14\ln(1+t^4)+\frac1{2\sqrt2}\ln\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}-\frac12\tan^{-1}t^2+c$$ where $t=\sqrt{\cot x}$
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Letting $x=\tan(u)$, $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{1+\tan(u)^2}\sec^2(u) \, du=u+k=\arctan(x)+k$$ Also,  $$\int\frac{1}{1+x^2} \, dx=\int\frac{1}{(x-i)(x+i)} \, dx=\frac{1}{2i}\int\frac{1}{(x-i)}-\frac{1}{(x+i)} \, dx$$ $$=\frac{1}{2i}\left(\ln(x-i)-\ln(x+i)\right)+c$$ $$=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+c$$ Giving $$\arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+q$$  Why is this correct? What is the nature of $q$ (is it 'flexible' so the equality doesn't really mean much)?  I think it probably has something to do with the relationship between $\log(z)$ and $\arg(z)$, but $\arg(z\pm i)$ is hard to calculate neatly.
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i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere.   $$\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$$  I've also tried letting $u = \frac{x^2+1}{x}$, $du/dx$ gets me $1-\frac{1}{x^2}$ but it doesn't seem to be working either.  Hope to receive some advise/ solutions on how to start tackling the question
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I'm learning about trig identities, and I'm struggling to prove that two expressions are equal:  $$ \frac{\sin(A + B)}{\sin(A - B)}=\frac{\tan A + \tan B}{\tan A - \tan B} $$  How do you go about proving this? I know about compound angles - i.e. the sine, cosine and tangent of $(A \pm B)$, but don't know how to apply it in this situation.
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If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$? Your help is appreciated.I tried various formulas but nothing is working.
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Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts.  below is his work.  How can I calculate $$ \int {\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x $$  My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$  Now Using Integration by Parts::  We get  $$= \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$$  Here he hit his road block.   I tried the opposite tactic   Taking the other approach by parts.   let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$    As you can see I got no further than he did.   So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way?
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Suppose $x \in (0,\frac{\pi}{2})$ and we want to prove $$x&lt;\frac{\sin x +\tan x}{2}$$I tried to prove it by taking $f(x)=\sin x+ \tan x -2x$ and show $f(x) &gt;0 ,when\space  x \in (0,\frac{\pi}{2})$ take f'$$f'=\cos x +1+\tan ^2 x-2\\=\tan^2 x-(1-\cos x)\\=\tan ^2 x-2sin^2(\frac x2)$$ I get stuck here ,because the last line need to be proved $\tan ^2 x&gt;2sin^2(\frac x2) ,when\space  x \in (0,\frac{\pi}{2})$ $\bf{Question}:$ Is there a geometrical method to prove the first inequality ? (or other idea)  Thanks in advance.    $\bf{Remark}: $I can see the function is increasing $when\space  x \in (0,\frac{\pi}{2})$ like below :
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My question is:  If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+\tan^2\frac{5\pi}{11}=Y$ then find $5X^2-Y$.  I couldn't find any way to simplify it. Please help. Thanks.
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My try, using $x = \sec(u)$ substitution:  $$ \begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &amp;=&amp; \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &amp;=&amp; \int \tan^2(u) \mathrm{d}u \\ &amp;=&amp; \tan(u) - u + C \\ &amp;=&amp; \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray} $$  However, according to , the answer should be: $$ \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C $$ When I derive this last answer I don't get back the integrand, but rather: $$ \frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)} $$  I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.  Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?