Question

square root 1+tan^2x/1-sin^2x all is under the square root.
i did sec^2theta/cos^2theta whats next?

Answer 1

sec^4 2x

Answer 2

you just need this trigonometric function to be solved or do you have anything to be proved?

Answer 3

find what it is equivalent to

Answer 4

then that's the answer! :)

Answer 5

\[\sqrt{1+\tan^2x}\div 1-\sin^2x\]

Answer 6

both of these r under the square root

Answer 7

ahh sorry i didn't notice that!then the answer is
sec^2 x

Answer 8

how did u get that?

Answer 9

1+tan^2 x= sec^2 x (that's trigonometry theory)
sin^x + cos^x=1 (that's trigonometry theory)
therefore [1-sin ^2 x= cos ^2 x]
so √1+tan2x ÷ √1−sin2x = √sec^2 x ÷ √cos ^2 x
=sec x ÷ cos x
=sec ^2 x
(because 1÷cos x =sec x)

Answer 10

oh i c i understand up to the 4th line and i got the answer on my own but up to the fourth line .wcan u explain line 5 and 6 for me .how did u cancel it and set up the fraction,.

Answer 11

=sec x ÷ cos x = sec x * 1/ cos x
= sec x * sec x ( because 1/cos x = sec x)
=sec ^2 x

Answer 12

k thanks so much and thanks for your patience .