Question

Proving trig equations

\(\sf 1+sec^2xsin^2x=sec^2x\)

Answer 1

I am thinking to use \(\sf 1+tan^2x=sec^x\) but I don't think that is right.

Answer 2

That should be \(\sf sec^2x\)

Answer 3

You're on right track.. keep going :)

Answer 4

Take the left hand side
\[1+\sec^2x\sin^2x\]

To use the identity, somehow you need to make the second term \(\sec^2x\sin^2x\) equal to \(\tan^2 x\). Maybe start by writing sec^2x as 1/cos^2x

Answer 5

Ok, \(\sf 1+\dfrac{1}{cos^2x}sin^2x=sec^2x\). I believe I somehow have to make \(\sf \dfrac{1}{cos^2x}sin^2x\) become \(\sf\dfrac{sin^2x}{cos^2x}\)

Answer 6

to multiply two fractions, you simply multiply the numerators and denominators separately :

\[\dfrac{1}{a}\cdot b = \dfrac{1}{a}\cdot \dfrac{b}{1} = \dfrac{1\cdot b}{a\cdot 1} = \dfrac{b}{a}\]

Answer 7

In short
\[\dfrac{1}{a}\cdot b = \dfrac{b}{a}\]

Answer 8

\(\large\color{black}{1+\sec^2x \sin^2x=\sec^2x}\)
\(\large\color{black}{\cos^2x\sec^2x+\sec^2x \sin^2x=\sec^2x}\)
\(\large\color{black}{(\cos^2x+ \sin^2x)\sec^2x=\sec^2x}\)

Answer 9

like this

Answer 10

ganesh, what do you think?:P

Answer 11

thats kinda cute :D

Answer 12

:D

Answer 13

Oh, I forgot that. So after multiplying there is \(\sf 1+\dfrac{sin^2x}{cos^2x}=sec^2x\).
\(\sf\dfrac{sin^2x}{cos^2x}=tan^2x\) Then \(\sf 1+tan^2x=sec^2x\) O_O

Answer 14

yuuuuuup, so simple, right?

Answer 15

oh, just rewriting it as tan^2x is definitely better

Answer 16

LoL

Answer 17

I haven't recognized that it is tan^2x sitting in there...