Question

@albhee14

Answer 1

\[\large \dfrac{ \tan^2A + 1}{\sec^2A }\]

Answer 2

is that the expression you're trying to simplify ?

Answer 3

yes :)

Answer 4

familiar with the identity : \(\sin^2A + \cos^2A = 1\) ?

Answer 5

yep

Answer 6

and we use another relation : \(\large \cos = \dfrac{1}{\sec }\)

Answer 7

\[\large \begin{align} \\ \dfrac{ \tan^2A + 1}{\sec^2A } = \large \cos^2A( \tan^2A +1) \end{align}\]

Answer 8

next, write `tan` as `sin/cos` and distribute ^^

Answer 9

ahhhh. okay.

Answer 10

\[\large \begin{align} \\ \dfrac{ \tan^2A + 1}{\sec^2A } &= \large \cos^2A( \tan^2A +1) \\
&= \large \cos^2A(\dfrac{\sin^2A}{\cos^2A} +1) \\

\end{align}\]

Answer 11

i hope you see how to work rest of the problem :)

Answer 12

so the answer will be cos^2A right? :D

Answer 13

hmm lets distribute and see what we get

Answer 14

\[\large \begin{align} \\ \dfrac{ \tan^2A + 1}{\sec^2A } &= \large \cos^2A( \tan^2A +1) \\
&= \large \cos^2A(\dfrac{\sin^2A}{\cos^2A} +1) \\
&= \large \cos^2A(\dfrac{\sin^2A}{\cos^2A}) +\cos^2A(1) \\
& = \large \sin^2A + \cos^2A \\
& = \large 1\\

\end{align}\]

Answer 15

see if that makes more or less sense...

Answer 16

ohh yah. I forgot about the 1

Answer 17

There are many methods to solve it:

First one is very direct, then use \(1 + tan^2(x) = sec^2(x)\)

Second method, Ganesh has already told you.

Third method is : Separating numerator parts and if you know:
\[\tan(x) = \frac{\sin(x)}{\cos(x)}\]

Answer 18

Thank you so much! :D

Answer 19

thank you also @waterineyes :D

Answer 20

But overall, Ganesh's explanation is always going to be best... :)

Answer 21

np :) you may ask your next question by clicking on "Ask a question" located the "top left" corner of this page... let me take a moment to welcome you officially

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Answer 22

:D thanks