Proving $\left(1-\cos^2x\right)\left(1+\tan^2x\right)=\tan^2x$

I need help proving this identity. I have tried simplifying it, but can't seem to get it to be true, although I have been told that it is. Can you help / sho all work?

$$\left(1-\cos^2x\right)\left(1+\tan^2x\right)=\tan^2x.$$

Edit: Here is the work I did (including errors):

So I know that $$(1-\cos^2 x) = \sin^2x$$

Then I changed $$\tan^2 x$$ to terms of $$(\frac{\sin x}{\cos x})^2$$.

Giving: $$(\sin^2 x) (1+ \frac{\cos^2x}{\sin^2x})$$.

From there I multiplied out getting: $$\sin^2x + (\sin^2 x+\frac{\cos^2x}{\sin^2x}).$$

Then simplifying from there I get $$\sin^2x+ \cos^2x$$.

• Welcome to Math.SE! You say that you've tried simplifying this; you should include that work (even if it's wrong) and explain where you got stuck. Information like this can help answerers to target their responses to your skill level, while avoiding telling you things that you already know. (Also, it helps avert suspicions that you're just trying to get people to do your homework for you.)
– Blue
Oct 3 '18 at 5:17
• Maybe you could show your effort how you simplify this and we might help you identify your problem or give you guidance
– xbh
Oct 3 '18 at 5:17
• So I know that (1-cos^2 x) = sin^2x Then I changed tan^2 x to terms of sin / cos. Giving: (sin^2 x) (1+ cos^2x/sin^2x) From there I multiplied out getting: sin^2x + ((sin^2 x)+(cos^2x))/(sin^2x) Then simplifying from there I get sin^2x+cos^2x Wait.. It worked this time I may have figured out where I went wrong. Oct 3 '18 at 5:20
• @CameronSkidmore Then I changed tan^2 x to terms of sin / cos And what did you get?
– dxiv
Oct 3 '18 at 5:21
• @CameronSkidmore: That's a good start! Please edit your question to include this work. (Not everyone reads the comments.) From there, it's mostly some symbol-juggling ... common denominators and stuff.
– Blue
Oct 3 '18 at 5:24

$$Proof \; that \; (1-\cos^2x)(1+\tan^2x)=\tan^2x \; for \; all \; positive \; reals \; x.$$

Proof:$$(1-\cos^2x)(1+\tan^2x)=\tan^2x.\tag{given claim}$$ $$\Downarrow\tag{factorising}$$ $$1+\tan^2x - \cos^2x - \cos^2x\tan^2x = \tan^2x.$$ $$\Downarrow\tag{subtracting \tan^2x from both sides}$$ $$1-\cos^2x-\cos^2x\tan^2x=0.$$ $$\Downarrow\tag{subtracting 1 from both sides}$$ $$-\cos^2x-\cos^2x\tan^2x=-1.$$ $$\Downarrow\tag{expanding the left hand side}$$ $$-\cos^2x(1+\tan^2x)=-1.$$ $$\Downarrow\tag{cancelling out the negatives}$$ $$\cos^2x(1+\tan^2x)=1.\tag1$$ Note that in a right angle triangle with sides $$a$$, $$b$$ and hypotenuse $$c$$, there exists an angle $$\theta$$ such that

\begin{align}\sin\theta &= a\div c \\ \cos\theta &= b\div c \\ \tan \theta &= a\div b = \frac{a\div c}{b\div c} = \frac{\sin\theta}{\cos\theta} \\ \\ \therefore \tan x &= \frac{\sin x}{\cos x}\tag*{(for all positive reals x) \; (2)}\end{align}

$$\Downarrow\tag{substituting (2) into Eq. (1)}$$ $$\cos^2x\left(1+\frac{\sin^2x}{\cos^2x}\right)=1$$ $$\Downarrow\tag{simplifying the right hand side}$$ $$\cos^2x+\sin^2x=1.\tag3$$ Since $$a^2+b^2=c^2$$ by the Pythagorean Theorem, and we have the equations in the sandbox, can you complete the proof? $$($$Hint: begin with Eq. $$(3)$$ and make substitutions.$$)$$

After showing how $$truth\Rightarrow claim$$, you can try showing how $$claim\Rightarrow truth$$, wherefore the proof is made stronger. (I assert this particular statement thanks to a user who looked at this answer. Thank you!)

• very well instructed proof.+1 Oct 3 '18 at 6:17
• @RakibulIslamPrince thanks! :D Oct 3 '18 at 6:17

L.H.S= $$(1-\cos^2 x)(1+\tan^2 x)$$ $$=\sin^2 x\cdot\sec^2 x$$ $$=\dfrac{\sin^2 x}{\cos^2 x}$$ $$=\tan^2 x=\text{R.H.S.}$$

• Thank you! This helped a lot. I can see that (1+tan^2 x) = sec^2 x. Knowing that sec = 1/cos, it makes it easy to multiply through and produce sin/cos! Oct 3 '18 at 5:34
• welcome.If it helps,then you should mark it as an answer.thank you! Oct 3 '18 at 5:36
• @CameronSkidmore yes, if it helps, mark it as an answer! $($It earnt my upvote, btw$)$ :D Oct 3 '18 at 21:39

Just mess around with the left hand side a bit.
$$(1-\cos^2 x)(1+\tan^2 x)$$ We know the following identity. $$1-\cos ^2 x = \sin^2 x$$ Now, simply replace $$1-\cos^2 x$$ with $$\sin^2 x$$. $$(\sin^2 x)\cdot(1+\tan^2 x)$$ $$\sin^2 x+\sin^2 x\cdot\tan^2 x$$ $$\sin^2 x+\sin^2 x\cdot\big(\frac{\sin^2 x}{\cos^2 x}\big)$$ $$\sin^2 x + \frac{\sin^4 x}{\cos^2 x}$$ Now, just multiply the numerator and denominator of the first fraction by $$\cos^2 x$$ so the two fractions can be added with a common denominator. $$\frac{\sin^2 x\cdot \cos^2 x}{\cos^2 x}+\frac{\sin^4 x}{\cos^2 x}$$ $$\frac{\sin^2 x\cdot \cos^2 x+\sin^4 x}{\cos^2 x}$$ Factor by $$\sin^2 x$$. $$\frac{\sin^2 x\cdot(\cos^2 x+\sin^2 x)}{\cos^2 x}$$ We also know the following identity. $$\sin^2 x+\cos^2 x = 1$$ So, just replace $$\sin^2x +\cos^2$$ with $$1$$. $$\frac{\sin^2 x\cdot(1)}{\cos^2 x} \implies \frac{\sin^2 x}{\cos^2 x} \implies \tan^2 x$$ Therefore, we have verified the identity. $$\boxed{(1-\cos^2 x)\cdot(1+\tan^2 x) = \tan^2 x}$$

• Shouldn't the identity be $(1-\color{red}{\cos^2x})\cdot (1+\tan^2x)=\tan^2x$? You wrote $\color{red}{\sin^2x}$ instead. Oct 3 '18 at 21:36
• Oh yes, wasn’t paying attention. Will edit it. Thanks for pointing it out. Oct 3 '18 at 21:49
• No problemo! Glad to help. $(+1)$ :D Oct 3 '18 at 21:51
• Elegant proof bro (+1) Oct 30 '18 at 17:31
• Thanks, although I'd personally say Rakibul's proof is the best. The simpler, the better! Oct 30 '18 at 17:43

Remember the idea sin²x +cos²x = 1, it yields 1-cos²x = sin²x

And sec²x - tan²x =1, it yields sec²x = 1+tan²x

By substituting both of these equations, you can easily proof the lefr hand side.

Can you finish it by using two equations above?

$$\dfrac{\tan^2x}{1+\tan^2x}=\dfrac{\sin^2x}{\cos^2x+\sin^2x}=1-?$$

Convert trig functions to Sines and cosines:

$$(1-\cos^2 x)(1+\frac{\sin^2 x}{\cos^2 x})=\frac{(1-\cos^2 x)(\cos^2 x+\sin^2 x)}{\cos^2 x}$$

and use $$\sin^2 x+\cos^2 x=1$$ to simplify the factors in the numerator.

HINT: $$1+tan^2x= 1+\frac{sin^2x}{cos^2x}=\frac{cos^2x}{cos^2x}+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}$$