Proving Trigonometric Functions

[Stardust1]

Hi guys, there was this one proof question for trigonometric functions that I came across and I can't prove it. I tried simplifying both sides and decomposing the cos^2(theta) into (1-sin^2(theta)) but its no use. the question is attached below. Can someone please show me how to solve this question?

 

[Harry_the_cat]

Rewrite both sides in terms of cos.

 

[R.M.]

Can you rewrite sin^2(x) into something involving cos^2(x)? If yes, clean up the numerator then put each term over the denominator and see if you recognize any other identities.

 

[fresh_42]

I find the motto: "remove what disturbs the most" a good advice that gets you started in many cases. Here it is the tangent that disturbs the most, but it can be replaced by the quotient of sine and cosine. Then we get an equation with two unknowns: $\sin \theta$ and $\cos \theta$ which are also related by Euler's formula, or Pythagoras if you like.

 

[Stardust1]

Thanks for your help guys! I was able to solve the question by rewriting both sides in terms of cos and then simplifying.

 

[Aion]

As suggested above, one can rewrite it in terms of cosine, then simplify it until you get $sin^2(x)+cos^2(x)=1$. But direct proof is simpler in my opinion.

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

 

[fresh_42]

As suggested above, one can rewrite it in terms of cosine, then simplify it until you get $sin^2(x)+cos^2(x)=1$. But direct proof is simpler in my opinion.

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

Indeed, the more as substitutions end up in $1=1 \text{ or }0=0$ and you have to write it down in this way anyway.
But the crucial points $(\tan=\sin / \cos \text{ and }\sin^2+\cos^2=1)$ remain the same.

 

[lookagain]

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

One more partial step should be included for the substitution of the identity for the
second fraction. The cancelation of terms leads directly to the final form.

$\displaystyle \dfrac{cos^2(\theta)cos^2(\theta)}{cos^2(\theta)}$