- sin q and cosq are base trigs
- sin2 q + cos2 q = 1
tan q = sinq
cosq
cot q = cosq
sinq
sec q = __1_
cosq
csc q = __1__
sinq
Prove the identity
1
- sin q = cot q cos q
sin q
Prove the identity
sin 2q =
tan q
1 +
cos2q
Prove the following
1. sec2x + cosec2x
= sec2x cosec2x
2. sec x cot x = cosec x
3. cos2x__ = 1 + sin x
1 – sin x
4. sin x + sin x cos2x
= cosec x
5. sin x cos x tan x = 1 –
cos2x
6. tan x + cot x = sec x
cosec x
7. 1 + cos x = sin x___
sin x 1 – cos x
1. let theta= X {}= explanation
ReplyDelete1/sin X - sin X = cot X . cos X
RHS
sin^2 X + Cos^2 X /sin X - sin X/1 {lcm found}
sin^2 X + Cos^2 X - sin^2 X / sin X {sin^2-Sin^2)
cos^2 X / Sin X {rewritten in a different exponent form}
Cos X/sin X * cos X/1 {cos/sin= cot, cos/1= cos}
cot X*cosX = CotX*CosX
1. let theta= x []= explanation
ReplyDelete1/sin x - sin x = cot x . cos x
lhs
cot x .cos x
cos x/sin x . cos x/1 [break cot into its roots]
cos^2 x/sin x [multiply cos.cos}
1 - sin^2 x/sin x
[from Pythagoras theorem sin^2 + cos^2 = 1
make cos^ 2the subject: cos^2= 1-sin^2 ,plug into expression]
1/sin x - sin^2x/sin x rewrite the expression
1/sin x - sin x = 1/sin x - sin x [ (sin^2x/sin x) cancels to make (sin x ) ]
1/sin x - sin x = 1/sin x - sin x <----- ans
sec x.cot x = cosec x
ReplyDelete1/cos [x] . 1/tan [x]
1/cos [x] . cos [x] / sin [x]
1/sin [x] = cosec x
2.
Delete5.
ReplyDeletesin x.cos x .tan x=1 - cos^2
lhs
sin x/1 . cos x/1 . sin x/cos x
sin^2 x/1
1- cos^2 x = 1-cos^2 x
if sec^2x+cosec^2x= sec^2xcosec^2x then how to start from on the right if there is no way to put any brackets. example: Cot^2 theta= cos^2 theta+(cot theta cos theta)^2.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteLHS
ReplyDeleteSince secx= 1/cosx and csec x = 1/sinx
then ---> sec^2 x+ cosec^2 x= 1/cos^2 x + 1/sin^2 x
= (sec^2 x + cos^2 x)/(sin^2 x)(cos^2 x)
=1/(sin^2 X*cos^2 X) `````sin^2 X+cos^2 X=1``````
=(sec^2 X)*(cosec^2 X)
=R.H.S