Question

Can someone explain on how to solve this?
sec^4 2theta?

Answer 1

what is sec 2 theta?

Answer 2

@student12345 you have got to answer me now :)

Answer 3

1 + tan^2 theta

Answer 4

@surjithayer , @student12345 doesn't know the 2theta's of sin , cos and tan , first ask @student12345 them :) :)

Answer 5

exactly, that's why i'm having a hard time. i've got to solve this, but no one discussed this to me since i wasn't present.

Answer 6

\[\sec ^{4}2\theta=\left( \sec ^{2}2\theta \right)^{2}=\left( 1+\tan ^{2}2\theta\right)^{^{2}}\]
\[=\left( 1+\left( \frac{ 2\tan \theta }{ 1-\tan ^{2}\theta } \right)^{2} \right)^{2}\]
\[=\left( \frac{ \left( 1-\tan ^{2}\theta \right)^{2}+4\tan ^{2}\theta }{ \left( 1- \tan ^{2} \theta\right)^{2} } \right)^{2} \]
try further

Answer 7

\[=\left( \frac{ 1+\tan ^{4}\theta-2\tan ^{2}\theta+4\tan ^{2}\theta }{\left( 1-\tan ^{2}\theta\right)^{2} } \right)^{2}\]
I don't know what you want ?

Answer 8

theta is what im finding.

Answer 9

What you mean?
Give clue.

Answer 10

either give me answer, so that I can understand.

Answer 11

i'm finding for theta. what is the angle.

Answer 12

give me complete statement of question, you are missing something.

Answer 13

for example,
\[2\sec \theta + 1 = \sec \theta\ +3\]
\[\sec \theta = 2\]
\[\theta = 60°\]

Answer 14

\[\sec ^{4}2\theta=?\]
give me equation

Answer 15

oh my bad , it's actually \[\sec^42\theta =4\]

Answer 16

\[\sin 2\theta=2\sin \theta \cos \theta ,\cos 2\theta=\cos ^{2}\theta-\sin ^{2}\theta \]
\[Also \sin2\theta =\frac{ 2\tan \theta }{ 1-\tan ^{2}\theta }\]

Answer 17

i dont get it lol

Answer 18

\[\sec ^{4}2\theta=4\]
\[\frac{ 1 }{ \cos ^{4}2\theta }=4\]
\[4\cos ^{4}2\theta=1\]
\[\left( 2\cos ^{2} 2\theta \right)^{2}=1\]
\[\left( 1+\cos 4 \theta \right)^{2}=1\]
\[1+\cos 4 \theta =\pm 1\]
\[Either1+\cos 4\theta=1\]
\[\cos 4\theta =0 =\cos n \frac{ \pi }{ 2},\theta =\frac{ n \pi }{ 8},where n is an integer\]
\[or \cos 4 \theta=-2,regected,\because \left| \cos 4 \theta \right|\le 1\]

Answer 19

If there is any doubt ask me.