Question

prove that tanA-cotA /sinAcosA =tan ^2A-cot^2A

Answer 1

Given, RHS = tan^2A - cot^2A

                   = (tanA-cotA)(tanA+cotA)

                   = (tanA-cotA)(sinA/cosA+cosA/sinA)

                   = (tanA-cotA)(sin^2A+cos^2A)/(SinA cosA) (sin^2a+cos^2a=1)

                   = (tanA-cotA)/sinAcosA

RHS = LHS.


Hope this helps!

Answer 2

Hi !

RHS = tan²A - cot²A

this can be written in the form of the identity (a-b) (a + b) = a² - b²

 =  ( tan A - cotA) (tanA + cotA)

we know that ,
tan A = sinA/cosA
cot A = cosA/sinA

i.e.,

= (tanA - cotA) (sinA/cosA + cosA/sinA)

(sinA/cosA + cosA/sinA) :- LCM = sinAcosA


= (tanA - cotA) ( sin²A + cos²A/sinAcosA)

but , we know that , sin²A + cos²A = 1 (identity)

so,

=  (tanA - cotA) ( 1/sinAcosA)

= (tanA - cotA)/sinAcosA

= LHS