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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. https://i.stack.imgur.com/v1eDl.jpg

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  • $\begingroup$ $\sin^2 x(1/\cos^2 x-1)=\sin^2 x \tan^2 x$, 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, 2014 at 12:37

4 Answers 4

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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.

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  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} $$
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For $\#1$ $$\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 $\#2$ $$\csc x=\frac1{\sin x},\sec x=\frac1{\cos x}$$

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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.

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