Oct 1, 2016 · 1 + tan2(x) = sec2(x) See explanation... Starting from: cos^2 (x) + sin^2 (x) = 1 Divide both sides by cos^2 (x) to get: cos^2 (x)/cos^2 (x) + sin^2 (x)/cos^2 (x) = 1/cos^2 (x) which simplifies to: 1+tan^2 (x) = sec^2 (x)

I need to prove that: $$1+\tan x \tan 2x = \sec 2x.$$ I started this by making sec 1/cos and using the double angle identity for that and it didn't work at all in any way ever. Not sure why I can't do that, but something was wrong.

tan^2 x = sec^2 x - 1 is an identity. We can prove the identity by using other trigonometric identities.

Jun 19, 2018 · This video explains the proof of 1+tan^2x=sec^2x using Pythagoras Theorem in the most simple and easy way possible. Please Share & SUBSCRIBE xoxo! Do you like Voice Explanation then...

Jan 9, 2018 · Yes, sec2 − 1 = tan2x is an identity. We start from sin2x + cos2x = 1. Then divide everything by cos2x. sin2x cos2x + cos2x cos2x = tan2x + 1 = 1 cos2x = sec2x. Rearrange to find tan2x. tan2x = sec2x −1.

Step 1: Simplify the numerator using the Pythagorean identity. We know that 1+tan2x=sec2x. Therefore, the expression becomes: Step 2: Simplify the fraction. Since the numerator and denominator are the same, the fraction simplifies to 1. The …

Because sin 2 x + cos 2 x = 1, we can simplify the numerator of the left hand side, meaning that (cos 2 x + sin 2 x)/cos 2 x = 1/cos 2 x which is sec 2 x (the right hand side). Therefore LHS=RHS and we have proven 1 + tan 2 x = sec 2 x. We know that tan …

Oct 30, 2021 · We have $\sin^2(x)+\cos^2(x)=1$. Divide both sides by $\cos^2(x)$ and rearrange to get your required result.

Mar 13, 2018 · If one is asked to prove $1+\tan^2(x)=\sec^2(x)$, this is how I would prove it. $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}=\sec^2(x)$$ and $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\frac{\sin^2(x)}{\cos^2(x)}=1+\left(\frac{\sin(x)}{\cos(x)}\right)^2=1 ...

Now, we use the basic trigonometric formula \[\sec x = \dfrac{{\left( {{\text{Hypotenuse}}} \right)}}{{\left( {{\text{Adjacent Side}}} \right)}}\], so we have, \[{\sec ^2}x = R.H.S.\] As $ L.H.S=R.H.S$, hence the given identity proved.