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

Question

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 $\sin^2x+ \cos^2x$.

Answer 1

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

Answer 2

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

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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!)

Answer 3

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?

Answer 4

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

Answer 5

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.

Answer 6

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. $$\boxed{(1-\cos^2 x)\cdot(1+\tan^2 x) = \tan^2 x}$$

Answer 7

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

Answer 8

You know $1-cos^2(x) = sin^2(x)$ and $1+tan^2(x)=sec^2(x)$. Hence,

$(1-cos^2x)(1+tan^2x)=sin^2(x)sec^2(x)$. But, $sec(x) = \frac{1}{cos(x)}$ .