$$\eqalign{\tan^2\theta-\sec^2\theta &=\tan^2\theta-\dfrac1{\cos^2\theta}\\&=\dfrac{\sin^2\theta}{\cos^2\theta}-\dfrac1{\cos^2\theta}\\&=-\dfrac{\cos^2\theta}{\cos^2\theta}\\&=-1.}$$ note:$\theta≠ (2k+1)\frac{\pi}{2}$, $k$ : integer number
May i know if my answer is correct or not..
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This is in general true apart from edge cases where $tan(\theta), sec(\theta)$ are undefined. What is nice about trig functions is that very often they can simplify nicely, so that complicated statements about circles and triangles can be solved using calculus, or calculus can be solved using trig rules. It is also common for equations to turn out to be constant after some other part of the problem is applied. What is interesting that this is not always an artifact of a problem "designed to be solved," but rather an artifact of the things we study. There are a great many problems in multi-variable calculus that dealt with very general figures that were made easier by applying trigonometric rules like these. If you plan on continuing in mathematics through and past calculus, this approach will be important, powerful, and relatively simple to apply.
You seem to have a penchant for expressing everything in terms of sines and cosines when simplifying trigonometric expressions. I used to do the same (because that's the way that Teacher did it), but I soon found it to be too cumbersome.
After a while, I started using 3 different triangles to help me out. Please note that I didn't draw the 3 triangles to scale. Also, I always expressed "$1$" as "$1^2$" to remind me that I'm using the Pythagorean theorem.
By using the second triangle we have $$\tan^2\theta - \sec^2\theta = (\sec^2\theta - 1^2) - \sec^2\theta = -1^2 = -1$$