Question

plsee prove that

Answer 1

Try to convert the expression on the right using these
\[\tan^2A=\frac{\sin^2A}{cos^2A}\]
\[\tan^2B=\frac{\sin^2B}{\cos^2B}\]

Answer 2

Sorry it is left side of the equation.

Answer 3

\[\tan^2A-\tan^2B\]
\[=\frac{\sin^2A}{\cos^2A}-\frac{\sin^2B}{\cos^2B}\]
well I guess the expression on the right should be this
\[\frac{\sin^2A-\sin^2B}{\cos^2A\cos^2B}\]

Answer 4

\[\tan^2A-\tan^2B\]
\[=\frac{\sin^2A}{\cos^2A}-\frac{\sin^2B}{\cos^2B}\]
\[=\frac{\sin^2A\cos^2B-\sin^2B\cos^2A}{\cos^2A\cos^2B}\]
\[=\frac{\sin^2A(1-\sin^2B)-\sin^2B(1-\sin^2A)}{\cos^2A\cos^2B}\]

Answer 5

Does that help? @msingh

Answer 6

hmm, i got it
@ajprincess
thanks

Answer 7

Maybe use the secant identity tan^2x= sec^2x -1 for both the tans

Answer 8

\[\Large \sec^2 A- 1 -(\sec^2B - 1) = \sec^2A-\sec^2B\]

Answer 9

\(\large \cfrac{1}{\cos^2 A } - \cfrac{1}{\cos^2 B} = \cfrac{\cos^2 B - \cos^2A}{\cos^2 A \cos^2 B} \)

\(\large \implies \cfrac{1-\sin^2 B - 1 + \sin^2 A}{\cos^2 A \cos^2 B } \)

\(\large \implies \cfrac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} \)

@agent0smith is right. This might be much easier.

Answer 10

ya I agree with u @mathslover. @agent0smith's method is much easier.

Answer 11

Yep but still, if the asker can understand your method much easily then it is better to go with that one. Though, I prefer , one should understand all the methods available :)

Answer 12

Still not sure how we get rid of that squared in the denominator, though...

Answer 13

ya. guess there is a mistake in the ques.:)

Answer 14

I tried putting it on wolframalpha and it doesn't appear to say the orig. question is false... http://www.wolframalpha.com/input/?i=tan%5E2A-tan%5E2B+%3D+%28sin%5E2A-sin%5E2B%29%2F%28cosA*cosB%29

Answer 15

http://www.wolframalpha.com/input/?i=%28sin%5E2A-sin%5E2B%29%2F%28cosA%5E2*cosB%5E2%29+%3D+%28sin%5E2A-sin%5E2B%29%2F%28cosA*cosB%29

It seems they're equivalent and I'm trying to figure out how

Answer 16

Oh i is cosA cosB in denominator, didn't notice that.

Answer 17

I don't think WA is actually comparing them, just solving them... i tried to get it to say if it's true but it didn't like that...

I feel like there is a mistake. I don't see how they can be equal.

Answer 18

I tried to see the alternate forms in which tan^2A-tan^2B be written. None of them match the given expression

Answer 19

Question is wrong or it is true for a special condition.

Answer 20

Yeah, i don't see how the two denominators are equal. cos^2Acos^2B is always positive. cosAcosB is not.

Answer 21

Yeah.

Answer 22

@agent0smith can you please tell me what is your way to do that problem?

Answer 23

The way @mathslover did it was the way i did it.

Answer 24

That was the only way that didn't lead to mess of sins and coses

Answer 25

can you again show your work for better understanding for me!!

Answer 26

@agent0smith can you?

Answer 27

@mayankdevnani try to understand it from above.

Answer 28

\[\large \cfrac{1}{\cos^2 A } - \cfrac{1}{\cos^2 B} = \]get common denominators here

Answer 29

i can't understand nicely. that's why i ask?

Answer 30

Mayank, what are you not getting in the above work?

Answer 31

\[\large \cfrac{1}{\cos^2 A } - \cfrac{1}{\cos^2 B} =\]
\[\large \frac{1}{\cos^2 A }*\frac{\cos^2 B}{\cos^2 B }-\frac{1}{\cos^2 B} *\frac{\cos^2 A}{\cos^2 A }\]

Answer 32

where does \[\frac{1}{\cos^2A}- \frac{1}{\cos^B}\] comes?

Answer 33

@agent0smith and @mathslover

Answer 34

hey!!!! guys!!! i completely understand ....thank you @agent0smith

Answer 35

i am not concentrate at this question because recently i ask a physic question!!!
SORRY FOR DISTURBING YOU!!!!

Answer 36

lol it's okay.

Answer 37

:)