how to prove $\tan A+\sec A=\frac{1}{(\sec A-\tan A)}$ ?
I already tried: $$\begin{align} \sin A/\cos A+1/\cos A&=1/(\sec A-\tan A)\\ \sin A+1/\cos A&=1/(\sec A-\tan A)\\ \end{align}$$
$\endgroup$ 47 Answers
$\begingroup$$\mathbf{HINT:}$$$ \begin{align} (\tan A+\sec A)(\sec A-\tan A)&=\require{cancel}\cancel{\tan A\sec A}-\tan^2 A+\sec^2 A\,\, \require{cancel}\cancel{\ -\tan A \sec A} \\ &=\sec^2 A -\tan^2 A\equiv1 \end{align}$$
$\endgroup$ $\begingroup$Hint: $\tan^2 \alpha + 1 = \sec^2 \alpha$.
$\endgroup$ $\begingroup$Sometimes we just need to stop the sin/cos madness. In many situations it is certainly appropriate to convert everything into sines and cosines, but in this problem it just adds unnecessary complexity.
First, analyze the problem as follows
$$\tan A+\sec A = \frac{1}{\sec A - \tan A}$$ $$(\tan A+\sec A)(\sec A - \tan A) = (\frac{1}{\sec A - \tan A})(\sec A - \tan A)$$ $$\sec^2A-\tan^2A = 1$$
But this is just a rearrangment of the familiar identity
$$\sec^2A=\tan^2A+1$$ The preceding suggests that we need to multiply top and bottom by $\sec A-\tan A$ $$\tan A+\sec A=(\tan A+\sec A)\frac{\sec A-\tan A}{\sec A-\tan A}=\dots$$
$\endgroup$ $\begingroup$$$\sin A+1/\cos A=1/(\sec A-\tan A)$$
Parentheses! And let's remember that we haven't shown this yet.
$$(\sin A+1)/\cos A=^? 1/(\sec A-\tan A)$$
Ahh, much better. Well, why stop there?
$$(\sin A+1/\cos A)=^? 1/((1/\cos A)-(\sin A / \cos A))$$
Well, get simplifying. It's impossible to fail to solve a problem; it's only possible to stop trying.
$\endgroup$ $\begingroup$HINT: $$\tan A+\sec A=\sin A/\cos A+1/\cos A=\frac{\sin A+1}{\cos A}\frac{1-\sin A}{1-\sin A}=\ldots$$ $$\ldots=\frac{\cos A}{1-\sin A}=\frac{\cos A/\cos A}{1/\cos A-\sin A/\cos A}$$
$\endgroup$ $\begingroup$See $\tan A$ as $\frac{\ sinA}{\cos A}$ and $\sec A$ as $\frac{1}{\cos A}$
1 part --------> $$\frac{\sin A+1}{ \cos A}$$ , 2 part---------> $$\frac{\cos A}{1-\sin A}$$
Focuse on the second part, multiply by $$\frac{1+\sin A}{1+\sin A}$$
So
$$\frac{\cos A}{1-\sin A}*\frac{1+\sin A}{1+\sin A}$$
Remembering that $\cos^2A=1-\sin^2A$ you will obtain
$$\frac{\sin A+1}{\cos A}=\frac{\sin A+1}{\cos A}$$
$\endgroup$ $\begingroup$$\sec ^2\alpha -\tan ^2\alpha=1$
$(\sec \alpha + \tan \alpha) (\sec \alpha - \tan \alpha)=1$
$\implies \sec \alpha + \tan \alpha =\frac{1}{\sec \alpha - \tan \alpha}$
$\endgroup$
