Question

Q9 pls ans fast ............

Answer 1

Your answer is ----

LHS = sinA/(1+cosA)

multiply numerator and denometnator by (1-cosA) , we get


sinA(1-cosA)/1-cos^2A

°•° 1-cos^2A = sin^2A

So, LHS = sinA(1-cosA)/sin^2A

=>LHS = (1-cosA)/sinA

divide numerator and denometnator by sinA

=> LHS = (1/sinA)-(cosA/sinA)/(sinA/sinA)

=> LHS = cosecA - cotA (°•° 1/sinA = cosecA and cosA/sinA = cotA)

=> LHS = cosecA-cotA = RHS


Hence proved

Answer 2

to prove: sinA/1+cosa=cosecA-cotA
taking l.h.s. ,sinA/1+cosa
multiply numerator and denominator by (1 -cosa)
so. sinA(1-cosa)/1+cosa(1-cosa)
= sinA(1-cosa)/1-cos^2a
using identity sin^2a+cos^2a=1
= sinA(1-cosa)/sin^2a
= 1/sina-cosa/sinA
coseca-cotA. = RHS