Question

establish the identity: [1+ sinø/1 - sinø] - [1 - sinø/1 + sinø] = 4tanøsecø

Answer 1

I've been trying the left hand side but I was wondering if the right side would be easier to attempt

Answer 2

both sides are easier to attempt madam.

Answer 3

\[\frac {1+ \sinø}{1 - \sinø} - \frac {1 - \sinø}{1 + \sinø} = 4\tanø\secø\]Its easier to simplify the left side than to complicate the right side.

Answer 4

\[ \frac {(1+ \sinø)^2-(1 - \sinø)^2}{(1 + \sinø)(1 - \sinø)} = 4 \tanø \sec ø\]\[\frac {4 \sin x}{1 - \sin^2 x} = 4 \tan x \sec x\]\[\frac {4 \sin x}{\cos^2 x} = 4 \tan x \sec x\]\[\frac {4\sin x}{\cos x} \times \frac {1}{\cos x} = 4 \tan x \sec x\]\[4 \tan x \sec x = 4 \tan x \sec x\]

Answer 5

I replaced the weird theta's w/ the x's since x's are easier to type.

Answer 6

For the first step, get both fractions to have the same denominators and combine them. After that, continue simplifying and substituting.

Answer 7

okay thank you! I get to the first step you did but i'm still confused how to get to the second step

Answer 8

Once you square both of the things and subtract, the numerator simplifies itself to that. The denominator is a difference of squares.

Answer 9

Oh okay I got it now. Thank you!

Answer 10

Your welcome! =)

Answer 11

of course LHS is easy but not too difficult RHS.. well good luck ... prove anything = anything. lol... heheh...