Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

(sin x/1-cos x) + (sin x/ 1+cos x)= 2csc x

Answer 1

@SithsAndGiggles @surjithayer

Answer 2

i think i figured it out

Answer 3

sin x/(1-cosx)=cot(x/2)
sin x/(1+cosx)=tan(x/2)
cot(x/2)+tan(x/2)=2 cscx
is this how to do it?

Answer 4

\[\frac{ \sin x }{ 1-\cos x }+\frac{ \sin x }{ 1+\cos x }=\sin x \left( \frac{ 1 }{ 1-\cos x } +\frac{ 1 }{ 1+\cos x }\right)\]
\[=\sin x \left( \frac{ 1+\cos x+1-\cos x }{ 1-\cos ^2x } \right)=\frac{ 2 \sin x }{ \sin ^2x }=\frac{ 2 }{ \sin x }=?\]

Answer 5

2/sinx= 2csc x

Answer 6

correct.

Answer 7

can you help me with one more? @surjithayer

Answer 8

i will try.

Answer 9

-tan^2x+sec^2x=1

Answer 10

we know \[\sin ^2x+\cos ^2x=1\]
divide both sides by \[\cos ^2x\]

Answer 11

i think now you can solve.

Answer 12

sin^2x=1/cos^2x

Answer 13

?

Answer 14

\[\frac{ \sin ^2x }{ \cos ^2x }+\frac{ \cos ^2x }{ \cos ^2x }=\frac{ 1 }{ \cos ^2x }\]
\[\tan ^2x+1=\sec ^2x, \]

Answer 15

how does that equal to one? @surjithayer

Answer 16

\[1=\sec ^2x-\tan ^2x\]

Answer 17

or \[-\tan ^2x+\sec ^2x=1\]

Answer 18

ooooh I see now!! sorry :p thank you!

Answer 19

yw