Question

(cosec x - sin x) (sec x - cos x) = 1 / tan x +cot x

Answer 1

yes

Answer 2

LHS: (1-sin^2x)/sinx * (1-cos^2x) /cosx
={cos^2xsin^2x}/sinxcosx
=sinxcosx

RHS: tanx/tan^2x+1
=tanx/sec^2x
=sinxcosx

Hence proved.

Answer 3

\(\large{( \csc x - \sin x)(\sec x - \cos x) = \cfrac{1}{\tan x + \cot x}}\)
[ RTP - Required to Prove]

\((\cfrac{1}{\sin x} - \sin x)( \cfrac{1}{\cos x} - \cos x )\)

\((\cfrac{1-\sin ^2 x}{\sin x} )(\cfrac{1- \cos^2 x}{\cos x})\)

Answer 4

Use : \(1 - \sin^2 x = \cos ^2 x\) and \(1- \cos ^2 x = \sin^2 x\) .

Answer 5

Just what I did, captain obvious. ^ -_-"

Answer 6

\(\large{\cfrac{\sin x \cos x}{1} = \cfrac{\sin x \cos x}{\sin^2 x + \cos^2 x}}\)
As : \(1 = \sin^2 x + \cos^2 x\)

\(\large{\cfrac{\cfrac{\sin x \cos x}{\sin x \cos x}}{\cfrac{\sin^2 x}{\sin x \cos x} + \cfrac{\cos ^2 x}{\sin x \cos x}}}\)

\(\large{\cfrac{1}{\cfrac{\cancel{\sin^2 x}^ {\sin x}}{\cancel{\sin x} \cos x} + \cfrac{\cancel{\cos^2 x }^{\cos x}}{\cancel{\cos x} \sin x}}}\)

\(\large{\cfrac{1}{\tan x + \cot x}}\)

Answer 7

@NeetziD now feel the difference. You solved the LHS first to get \(\sin x \cos x\) and then solved RHS to get the same : \(\sin x \cos x\) . Sorry, but I consider that as verifying not as proof , it is a proof but not a perfect one. There may be some other opinions regarding my thinking. Though, your method is 99% correct according to me.
I just presented a method where you can use LHS to find RHS .

Answer 8

agree ^^

Answer 9

@kewal I hope you got it. \( \begin{array}{}
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Answer 10

get it... =)

Answer 11

@mathsolver very much thnx for yur help, r u math teacher..?

Answer 12

You're welcome Kewal and no, I am not a math teacher. Just a student like you :)

Answer 13

i jst came 2 tenth grade this year and was having problems with trigo.. thanx again fr your help.

Answer 14

Good, keep it up!