Question

16. cos x (tan x + 2) (2 tan x + 1) = 2 sec x +5 sin x​

Answer 1

Answer:

LS = cosx(2tan^2 x + 5tanx + 2)

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx= (2 + 5sinxcosx)/cosx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx= (2 + 5sinxcosx)/cosx= 2/cosx + 5sinx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx= (2 + 5sinxcosx)/cosx= 2/cosx + 5sinx= 2secx + 5sinx

LS = cosx(2tan^2 x + 5tanx + 2)= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx= 2sin^2 x/cosx + 5sinx + 2cosx= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx= (2 + 5sinxcosx)/cosx= 2/cosx + 5sinx= 2secx + 5sinx= RS.