Question

Prove/verify
SinB + cosBcotB = cscB

Answer 1

I got disconnected.

Answer 2

ah okay.

Answer 3

\(\large\color{black}{ \bf SinB + cosBcotB = cscB }\)

\(\large\color{black}{ \bf SinB + cosB(cosB/sinB) = cscB }\)

\(\large\color{black}{ \bf SinB (SinB/SinB)+ cosB(cosB/sinB) = cscB }\)

read this for now, I am going to post more...

Answer 4

\(\large\color{black}{ \bf (~Sin^2B/SinB~)~+(~ cos^2B/sinB~) = cscB }\)

\(\large\color{black}{ \bf (~~ Sin^2B+cos^2B~~)/SinB = cscB }\)

\(\large\color{black}{ \bf 1/SinB = cscB }\)

\(\large\color{black}{ \bf cscB = cscB }\)

Answer 5

there :)

Answer 6

thank you im still trying to process this

Answer 7

I am going to post a couple rules that I used.


\(\large\color{blue}{ 1)~~~~\bf sin^2B+cos^2B=1 }\)

\(\large\color{blue}{ 2)~~~~\bf 1/sin^2B=csc^2B }\)

\(\large\color{red}{ 3)~~~~\bf cot(B)=cosB/sinB }\) and alternatively,
\(\large\color{red}{ 3)~~~~\bf cot^2(B)=cos^2B/sin^2B }\)

the rest is algebra.

I gtg bye :)

Answer 8

thanks

Answer 9

anytime !