Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

PROVING IDENTITIES!!

OpenStudy (anonymous):

qn?

OpenStudy (anonymous):

the answer is a calculator

OpenStudy (anonymous):

\[(\sin x + \cos x)(\tan x + \cot x) =\sec x + \csc x\]

OpenStudy (mertsj):

Change to sin and cos then FOIL it out

OpenStudy (anonymous):

Ok. We have - (sin x + cos x ) (sinx/cosx + cosx/sinx) (since tan x = sinx/cosx and cot x = cosx/sinx) => (sin x + cos x) (sin^2x + cos^2x/ sin x . cos x) => (sin x + cos x) ( 1/sin x. cos x ) (since sin^2x + cos^2x = 1) => (sin x + cos x)/sin x. cos.x Separating sin x, cos x in numerator with common denominator, sin x / sin x. cos x + cos x . / sin x cos x => 1/cos x + 1/sin x => sec x + csc x (since 1/cos x = sec x, 1/sin x = csc x) Proved

OpenStudy (anonymous):

\[(\sin x + \cos x)(\frac{ \sin x }{ \cos x }+\frac{ \cos x }{ \sin x })\] \[(\sin x + \cos x)(\frac{ \sin ^2 x + \cos ^2 x}{ \cos x \sin x })\] \[(\sin x + \cos x)(\frac{ 1 }{ \cos x \sin x })\]

OpenStudy (anonymous):

i got it....thank you all...

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!