Question

how to do this?

Answer 1

we should start from left side

Answer 2

ok

Answer 3

\[sinx(cotx+cosx.tanx)\\here~u~have~\to~use~two~basic~identities\\tanx=\frac{sinx}{cosx}\\and cotx=\frac{ 1 }{ tanx }\]

Answer 4

ok

Answer 5

so it should be \[sinx.cotx+sinx.cosx.tanx=\frac{ sinx.cosx }{ sinx }+\frac{ sinx.cosx.sinx }{ cosx }\\=cosx+\sin^2x\]
so its false

Answer 6

=cosx+sin2x but that part is right? can you explain me why it should be false? i am sorry i am really dumb in math

Answer 7

because we are converting left side to right side ,thats why its false
and
in the first step i just multiplied sinx which was outside the brakets
then just used
cotx=cosx/sinx
and
tanx=sinx/cosx

Answer 8

ohh ok i see now...

Answer 9

:)

Answer 10

can you help with this one too?

Answer 11

i can but i have to go offline for some 30 minutes ,sorry :(
u can try what is says i.e first cross-multiply

Answer 12

oh ok... well thank you so much for your help sir! also thank you for explaining and being patient! @sidsiddhartha