Question

Simplify :
sec^2 (theta) x cos^2 (theta) + tan^2 (theta)

Answer 1

what

Answer 2

post the problem

Answer 3

\[\sec^2 \theta = \dfrac{1}{\cos^2 \theta}\]

Answer 4

\[\sec ^{2}\theta \times \cos ^{2}\theta + \tan ^{2}\]

Answer 5

we should simplify it ... !!

Answer 6

Is it\[\sec^2 \theta (\cos^2 \theta + \tan^2 \theta)\]Or\[(\sec^2\theta\cdot \cos^2 \theta) + \tan^2 \theta\]

Answer 7

the second expression

Answer 8

Oh, not sure what the big deal with that is then. \[\sec^2 \theta = \dfrac{1}{\cos^2\theta}\]Therefore,\[\sec^2 \theta \cdot \cos^2\theta = \dfrac{1}{\cos^2\theta}\cdot \cos^2 \theta\]

Answer 9

it's not complete
\[\frac{ 1 }{ \cos ^{2}\theta } \times \cos ^{2} \theta + \frac{ \sin ^{2} \theta}{ \cos ^{2}\theta }\]

then what ??

Answer 10

You can start by cancelling out \(\cos^2 \theta\)...

Answer 11

\[\dfrac{1}{\cancel{\cos^2\theta}}\cdot \cancel {\cos^2 \theta }= 1\]

Answer 12

So you're now left with \(1 + \tan^2 \theta\). Does that remind you of an identity?

Answer 13

So ...
\[\frac{ 1 }{ 1 } + \frac{ \sin ^{2}\theta }{ \cos ^{2} }\]

then ?

Answer 14

\[1 + \dfrac{\sin^2\theta}{\cos^2 \theta}\]\[=\dfrac{\cos^2\theta}{\cos^2\theta} + \dfrac{\sin^2\theta}{\cos^2\theta}\]\[= \dfrac{\overbrace{\cos^2 \theta + \sin^2 \theta}^{1}}{\cos^2 \theta}\]\[= \dfrac{1}{\cos^2\theta}\]\[= \sec^2 \theta\]

Answer 15

Oh , Thank you a lot

Answer 16

There's also a straightforward identity that \(1 + \tan^2 \theta = \sec^2 \theta\) and there's another which says that \(1 + \cot^2 \theta = \csc^2 \theta\). Keep these two in mind.

Answer 17

I'm glade that you helped me
Thanks a lot

Answer 18

No problem :)