Question

trigonometric equations:-
(1-tan@)(1+sin2@)=1+tan@
@=theta

Answer 1

sin2 @ is sin squared or just sin2 ?

Answer 2

simply sin2@...not squared...
=2sin@cos@

Answer 3

do you know it sure that

tan x = (sin x)/cos x

yes ?

Answer 4

yup..

Answer 5

try rewriting this equation subtituting tan x = (sin x)/cos x

because you need getting an equation with just one the same unknowed term with theta

for example sin theta or cos theta

ok?

Answer 6

(cos@-sin@ / cos@)(sin2@)=(cos@+sin@ / cos@)

Answer 7

yes nice but sin2@ rewrite it how you have wrote in your first answer

Answer 8

(cos@-sin@ / cos@)(1+2sin@cos@)=(cos@+sin@ / cos@)

Answer 9

how you have got this 1 from 1+2sinxcosx ?

Answer 10

first you have wrote sin2x = 2sinxcosx

Answer 11

its already in the ques...i forgot to write it

Answer 12

ok than rewrite it please the correct exercise

Answer 13

its is there 1+sin2@

Answer 14

yes i see it

Answer 15

so if you check it again you see that there are two fractions with the same denominator cos x

Answer 16

cos@

sorry

Answer 17

yes ?

Answer 18

(cos@-sin@ / cos@)(1+2sin@cos@)=(cos@+sin@ / cos@)

(cos@ -sin@) (cos@ +sin@)
----------- * (1 +2sin@cos@) = ------------
cos@ cos@

so now do you understand it ?

Answer 19

so and hope you know again sure that sin^2 @ +cos^2 @ =1

yes ?

Answer 20

than on the left side inside paranthese there is an 1 what you can rewrite it like sin^2 @ +cos^2 @

ok ?

Answer 21

so and in this case you will get the formula a^2 +2ab +b^2 what you know that is equal
(a+b)^2

Answer 22

do you understand it ?

Answer 23

so than will get

(cos@ -sin@)(cos@ +sin@)^2 =(cos@ +sin@)

divide both sides by (cos@ +sin@) and will get

(cos@ -sin@)(cos@ +sin@)=1

but you see that on the left side there is formula (a+b)(a-b)=a^2 -b^2