Question

prove that (secx+cscx)/(tanx+cotx)=sinx+cosx

Answer 1

\(\sec x = \cfrac{1}{\cos x}\)
\(\csc x = \cfrac{1}{\sin x}\)
\(\tan x = \cfrac{\sin x}{\cos x}\)
\(\cot x = \cfrac{\cos x}{\sin x}\)

Use these values and put them in the equation

Answer 2

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

Take the LCM and simplify

Answer 3

@tinasaurusrex can you tell me what you get after simplifying it!

Answer 4

this is how far i got before i got stuck

Answer 5

Ok let us solve for numerator first :
\(\cfrac{1}{\cos x} + \cfrac{1}{sin x}\)

Can you take LCM and add them?

Answer 6

hint : \(\cfrac{1}{a} + \cfrac{1}{b} = \cfrac{b + a}{ab}\)

Answer 7

so you would have sinx+cosx/sinxcosx

Answer 8

Yeah! Similarly solve denominator..

Answer 9

\(\cfrac{\sin x}{\cos x} + \cfrac{\cos x}{\sin x}\) = ?

Answer 10

sinxcosx+cosxsinx/sinxcosx ? this one kind stumps me

Answer 11

No..

\(\cfrac{b}{c} + \cfrac{d}{e} = \cfrac{be + dc}{ce}\)

Answer 12

so sin^2x + cos^2x/sinxcosx ?

Answer 13

yes so what is \( sin^2 x + cos^2 x \)?

Answer 14

1

Answer 15

so we have \(\cfrac{1}{\sin x \cos x}\)

Answer 16

then would i just invert and multiply?

Answer 17

yes

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

Answer 18

\(\cfrac{\sin x + \cos x }{1} \times \cfrac{\cancel{\sin x \ cos x}}{\cancel {\sin x \cos x}}\)

Proved!!!!!!!!

Answer 19

awesome!! thanks for the help