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${\sin^2(x)\over \tan^2(x)}$

I did this and then got stuck. Could someone give me some hints please?

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  • $\begingroup$ Wait . What did you do? And from wher did you get this question? $\endgroup$ – Qwerty Nov 29 '16 at 19:11
  • $\begingroup$ Convert the $\tan x$ function to $\frac {\sin x}{\cos x}$ when in doubt always convert to the more basic $\sin x$ and $\cos x$ functions. $\endgroup$ – Doug M Nov 29 '16 at 19:16
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Good job at arriving at $${1-\cos^2(x)\over \sec^2(x)-1}= {\sin^2(x)\over \tan^2(x)}$$ We know that $\tan x = \dfrac {\sin x}{\cos x}.$ So.... $${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$

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For the equation: $$\frac{1-\cos^2x}{\sec^2x-1}$$

Multiply the numerator and denominator by $\cos^2x$

We now get: $$\frac{\cos^2x-\cos^4x}{1-\cos^2x}$$

Separate this into two fractions: $$\frac{\cos^2x}{\sin^2x} - \frac{\cos^4x}{\sin^2x}$$

This can then be converted to: $$\cot^2x - \cot^2x\cos^2x$$

We take $\cot^2x$ common, and get: $\cot^2x\sin^2x$

When this is multiplied, this gives us: $\cos^2x$, or rather, $1-\sin^2x$

Hence proved.

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  • $\begingroup$ Quick formatting lesson: write \cos, \sin, \cot, \tan... to format $\cos, \sin, \cot, \tan...$ For example, not the difference between $tan x$ =$ tan x$, versus, $\tan x$ = $\tan x$ $\endgroup$ – amWhy Nov 29 '16 at 19:35
  • $\begingroup$ I'm new to this so I'm not fully aware of the workings. Thanks for the tip! $\endgroup$ – Dhruv Bansal Nov 29 '16 at 19:41
  • $\begingroup$ It takes time to learn and get up-and-running with mathjax. Pretty much ever function/operator is preceded by a backslash: \ln, \det, \sin, \cos, \gcd etc, yields $\ln, \det,\sin, \cos, \gcd$, etc.$ $\endgroup$ – amWhy Nov 29 '16 at 19:50
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Notice that $\sec^2(x)=\frac{1}{\cos^2(x)}$ Putting it in our equation on R.H.S. it becomes: $${1-\cos^2(x)\over \sec^2(x)-1}={1-\cos^2(x)\over\frac{1}{\cos^2(x)}-1}$$ $$=\frac{(1-\cos^2)(\cos^2(x))}{1-\cos^2(x)}$$ $$=\cos^2(x)=1-\sin^2(x)$$

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  • $\begingroup$ Don't give up on our conversation in Constructive Feedback; I tried to address TheGreatDuck's interruption. Things are great there when he's not on a big ego trip! :-) $\endgroup$ – amWhy Nov 29 '16 at 20:30
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we have $$\frac{1}{\sec(x)^2-1}=\frac{\cos(x)^2}{1-\cos(x)^2}$$ and from both we get $$\cos(x)^2=1-\sin(x)^2$$

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  • $\begingroup$ $$\frac{1}{\sec^2(x)-1}\neq \frac{1-\cos^2(x)}{\cos^2(x)}$$ $\endgroup$ – teadawg1337 Nov 29 '16 at 19:27
  • $\begingroup$ it looks good now. $\endgroup$ – Vidyanshu Mishra Nov 29 '16 at 19:40
  • $\begingroup$ yes it was a typo $\endgroup$ – Dr. Sonnhard Graubner Nov 29 '16 at 19:42

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