1
$\begingroup$

Stuck on two questions.

First one: $\tan^2x - \sin^2x = (\sin^2 x)(\tan^2 x)$

Tried solving that with the right side but wasn't able to.

Second one: $\csc x / \sec x = \cot x$

I tried solving this one with the left side doing,

L.S = $\csc x / \sec x $
L.S = $(1 / \sin x) / (1 / \cos x) $
L.S = $\sin x / \cos x\quad$ [(1's cancel out]
L.S = $\tan x$

Don't know what I'm doing wrong here.

For reference, here is the trig identity worksheet we're supposed to use. http://imgur.com/TXxaYUQ

$\endgroup$
  • $\begingroup$ $\sin^2 x(1/\cos^2 x-1)=\sin^2 x \tan^2 x$, just carrying $\sin^2 x$ and using a known expression for $1/\cos^2 x$ (which in turn comes from dividing $\cos^2 x + \sin^2 x =1$ by $\cos^2 x$). $\endgroup$ – orion May 12 '14 at 12:37
1
$\begingroup$

Note first that IF $\,\sin(x)= 0,\,$ then clearly also $\tan(x) = 0$ in which case the equation evaluates to $\,0 = 0\,$ which certainly is true.

Now, in the case that $\,\sin x\neq 0\,$ (so also $\tan x\neq 0$), we can divide both sides of the equation $$\tan^2 x - \sin^2 x = \sin^2 x\tan^2 x$$ by $\sin^2 x\tan^2 x,\,$ giving us $$\csc^2 x - \cot^2x = 1$$

But $\csc^2 x = 1 + \cot^2 x,\,$ so that gives us $$1 + \cot^2 x - \cot^2 x = 1,$$ as desired.

$\endgroup$
1
$\begingroup$
  1. $$ \begin{align} \tan^2x - \sin^2x&=\frac{\sin^2x}{\cos^2x}-\sin^2x\quad;\ \text{where}\ \tan x=\frac{\sin x}{\cos x}\\ &=\sin^2x\left(\frac{1}{\cos^2x}-1\right)\\ &=\sin^2x\left(\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)\\ &=\frac{\sin^2x}{\cos^2x}\left(1-\cos^2x\right)\\ &=\color{blue}{\tan^2x \cdot \sin^2x}\quad;\ \text{where}\ \sin^2x + \cos^2x=1. \end{align} $$ $$$$
  2. $$ \begin{align} \frac{\csc x}{\sec x}&=\dfrac{\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}}\\ &=\frac{1}{\color{red}{\sin x}}\times\frac{\color{red}{\cos x}}{1}\\ &=\frac{\cos x}{\sin x}\\ &=\color{blue}{\cot x}. \end{align} $$
$\endgroup$
0
$\begingroup$

For $\#1$ $$\sin^2x+\sin^2x\cdot\tan^2x=\sin^2x(1+\tan^2x)=\frac{\sin^2x}{\cos^2x}$$

$$\implies \sin^2x+\sin^2x\cdot\tan^2x=\tan^2x$$

For $\#2$ $$\csc x=\frac1{\sin x},\sec x=\frac1{\cos x}$$

$\endgroup$
0
$\begingroup$

I just wanted to point out that when doing proofs like this that your argument should lead from the initial form (e.g. $\tan ^2 x - \sin^2 x$) to the final form ($\sin^2x \tan^2 x$). That should be your final answer, but sometimes it is just easier to prove that the LHS and RHS are equal.

For example, my doodles may look like this $$\tan ^2 x - \sin^2 x = \sin^2x \tan^2 x$$ $$\tan ^2 x = \sin^2x \tan^2 x + \sin^2 x$$ $$\tan ^2 x = \sin^2x(\tan^2 x + 1)$$ $$\tan ^2 x = \sin^2x(\sec^2 x)$$ $$\tan ^2 x = \sin^2x\bigg(\frac{1}{\cos^2 x}\bigg)$$ $$\tan ^2 x = \tan ^2 x$$

Now, I'm going to use my doodles to present my real argument (the one I hand into my teacher). Recall that I ended up $\tan^2 x$, but started with $\tan ^2 x - \sin^2 x$, so I'm just going to take $\sin^2 x$ along for the ride for a bit, and work backwards from my doodles. $$\tan ^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x}-\sin^2 x = \sin^2 x \sec^2 x-\sin^2 x = \sin^2 x(\sec^2 x-1) = \sin^2 x \tan^2 x $$

Try doing the same for the other problem.

By the way, this is how I remember the Pythagorean identities. enter image description here

It's unfortunate that Geometry has been downplayed so much in our high schools in recent years.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.