Question

Proving identities.
Csc0/cot0 - cos0 = 1+ sin0/sin^30

Answer 1

change cosec and cot is terms of sin and cos ..in LHS

Answer 2

can u write the problem in the equation editor .?

Answer 3

can you show me step by step how to solve this?

Answer 4

please first ..present that question using equation editor..
the RHS part is not clear to me .

Answer 5

its 1+sinu/sin^3u

Answer 6

sin^3u is divided by 1+sinu or only sinu .??

Answer 7

1+sinu is devided by sin^3u

Answer 8

my equation editor is not working idk why sorry

Answer 9

even mine is not working too .!
change cosec as 1/sin ...and sot as cos/sin den put and simplify

Answer 10

1/sinu is devided by cosu-cosu/sinu ???

Answer 11

1/sinu divided by cosu/sinu - cosu

Answer 12

then?

Answer 13

den (1/cosu) -cosu
sin^2u/cosu

Answer 14

is it devided by sin^2u/cosu??

Answer 15

yes

Answer 16

is that all?

Answer 17

no ..wondering what to do next ..to arrive at RHS

Answer 18

hello i am awesome

Answer 19

\[\frac{Cscθ }{Cotθ} - Cosθ = \frac{1+ Sinθ }{Sin^3θ}\] this is what you meant, here follow me,\[\huge\color{blue}{ \frac{Cscθ }{Cotθ} - Cosθ = \frac{1+ Sinθ }{Sin^3θ} } \]
\[\huge\color{blue}{ \frac{ ~~~~\frac{1 }{Sinθ} ~~~~}{ ~~~\frac{Cosθ }{Sinθ}~~ } ~~ - Cosθ = \frac{1+ Sinθ }{Sin^3θ} } \]
\[\huge\color{blue}{ \frac{1 }{Sinθ} \div \frac{Cosθ }{Sinθ} - Cosθ = \frac{1+ Sinθ }{Sin^3θ} } \]
\[\huge\color{blue}{ \frac{1 }{Sinθ} \times \frac{Sinθ }{Cosθ} - Cosθ = \frac{1+ Sinθ }{Sin^3θ} } \]
\[\huge\color{blue}{ \frac{1 }{Cosθ} - Cosθ = \frac{1+ Sinθ }{Sin^3θ} } \]

Answer 20

\[\huge\color{blue}{ \frac{1}{Cosθ} -Cosθ =\frac{1+Sinθ}{Sin^3θ}} \]
\[\huge\color{blue}{ \frac{1}{Cosθ} - \frac{Cos^2θ}{Cosθ} =\frac{1+Sinθ}{Sin^3θ}} \]
\[\huge\color{blue}{ \frac{1-Cos^2θ}{Cosθ} =\frac{1+Sinθ}{Sin^3θ}} \]
\[\huge\color{blue}{ \frac{Sin^2θ}{Cosθ} =\frac{1+Sinθ}{Sin^3θ}} \]

Answer 21

\[\huge\color{blue}{ Sin^5θ =(1+Sinθ)Cosθ} \]\[\huge\color{blue}{ Sin^5θ =Cosθ+SinθCosθ} \]

Answer 22

too long