Question

Prove the identity
http://i.imgur.com/nRmKT0P.png
I'm really bad at these... If someone could help guide me through this and give me some tips it would be greatly appreciated. Thanks!

Answer 1

take left hand side,
rewrite tan as sin/cos,
factor out sin, and u will see wat to do next

Answer 2

take tan= sin/cos, take out sin common and then you will have (1/cos^2 -1) simplify it and you will get sin^2(sin^2/cos^2) and sin/cos =tan

Answer 3

How exactly do I factor out sin?

Answer 4

left hand side :-

\(\large \tan^2 - \sin^2\)

\(\large \frac{\sin^2}{\cos^2} - \sin^2\)

Answer 5

do u see \(\sin^2\) in both terms ?
pull \(\sin^2\) out

Answer 6

And then:
\[\frac{ \sin ^{2} }{ \cos ^{2} } - (1-\cos ^{2})\]?

Answer 7

dont touch sin^2 in second term

Answer 8

left hand side :-

\(\large \tan^2 - \sin^2\)

\(\large \frac{\sin^2}{\cos^2} - \sin^2\)

\(\large \sin^2 \left( \frac{1}{\cos^2} - 1 \right) \)

Answer 9

left hand side :-

\(\large \tan^2 - \sin^2\)

\(\large \frac{\sin^2}{\cos^2} - \sin^2\)

\(\large \sin^2 \left( \frac{1}{\cos^2} - 1 \right) \)

\(\large \sin^2 \left( \sec^2 - 1 \right) \)

Answer 10

I've never learnt sec before... and how did you get the third term I'm not that sure

Answer 11

sec = 1/cos

Answer 12

other two are :-

cosec = 1/sin

cot = 1/tan

Answer 13

Oh ok

Answer 14

also we have sec^2-1 = tan^2

Answer 15

left hand side :-

\(\large \tan^2 - \sin^2\)

\(\large \frac{\sin^2}{\cos^2} - \sin^2\)

\(\large \sin^2 \left( \frac{1}{\cos^2} - 1 \right) \)

\(\large \sin^2 \left( \sec^2 - 1 \right) \)

\(\large \sin^2 \left( \tan^2 \right) \)

Answer 16

I get all the working apart from the third step

Answer 17

How can s^2/c^2 - s^2 become s^2(1/c^2 -1)?

Answer 18

we need to memorize below :-
sin^2+cos^2 = 1
sec^2-tan^2 = 1
cosec^2-cot^2 = 1

Answer 19

ohk that step, how we pulled out s^2 ?

Answer 20

thats throwing u off ?

Answer 21

Yeah i'm not that sure :/

Answer 22

let me ask u a question,

take below expression :-
a + ab

Answer 23

can we write it as,

a(1 + b)

?

Answer 24

Yes

Answer 25

same way, we're taking out s^2 below :-

s^2/c^2 - s^2

s^2(1/c^2 - 1)

Answer 26

cuz, s^2 is there in both terms

Answer 27

Yes i see and then we make that sec^2-1 right? and use the rule where
sec^2-1=tan^2?

Answer 28

exactly !

Answer 29

Now I understand thanks so much :D

Answer 30

np :)