Hi could you please help me with 1+cot^2x/sec^2x ?

Question

7041 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T01:01:41

The expression simplifies to ${cot}^{2} x$ $cot^2x$ .

First, we need these basic trig identities:

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

$sec x = \frac{1}{cos x}$ $secx=1/cosx$

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

Now, using the Pythagorean identity,

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

we can derive a new identity if we divide everything by ${sin}^{2} x$ $sin^2x$ :

$\frac{{sin}^{2} x}{{sin}^{2} x} + \frac{{cos}^{2} x}{{sin}^{2} x} = \frac{1}{{sin}^{2} x}$ $sin^2x/sin^2x+cos^2x/sin^2x=1/sin^2x$

$1 + {cot}^{2} x = {csc}^{2} x$ $1+cot^2x=csc^2x$

Use all these to simplify our expression:

$= \frac{1 + {cot}^{2} x}{{sec}^{2} x}$ $color(white)=(1+cot^2x)/sec^2x$

$= \frac{{csc}^{2} x}{{sec}^{2} x}$ $=csc^2x/sec^2x$

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

$= \frac{1}{{sin}^{2} x} \cdot \frac{{cos}^{2} x}{1}$ $=1/sin^2x*cos^2x/1$

$= \frac{{cos}^{2} x}{{sin}^{2} x}$ $=cos^2x/sin^2x$

$= {cot}^{2} x$ $=cot^2x$

That's it. Hope this helped!