Question

(1+tanθ)/(sinθ+cosθ)=secθ

Answer 1

what to do?

Answer 2

\[\frac{1+\tan \theta}{\sin \theta+\cos \theta}=\sec \theta\]

Answer 3

verify indentity

Answer 4

(1+tanθ)/(sinθ+cosθ) = 1/cosθ
cosθ/(sinθ +cosθ) = 1/(1+tanθ)
tanθ = sinθ/cosθ
1 + tanθ = 1 + sinθ/cosθ = (cosθ + sinθ)/cosθ
1/(1+tanθ) = cosθ/(cosθ + sinθ)
which is what is in the left side :)

Answer 5

put any value of theta and you will se that the rhs and lhs are same

Answer 6

RHS\[\sec \theta=\frac{\sin \theta + \cos \theta}{\sin \theta +\cos \theta}\sec \theta\]\[=\frac{1+\tan \theta}{\sin \theta + \cos \theta}=RHS\]

Answer 7

you could actually do it without my first steps .. :
1 + tanθ = 1 + sinθ/cosθ = (cosθ + sinθ)/cosθ
so put it in the original :
( (cosθ + sinθ)/cosθ) / ( sinθ + cosθ) =1 /cosθ
( (cosθ + sinθ)/(cosθ( sinθ + cosθ))) = 1/cosθ
then 1/cosθ = 1/cosθ
whatever you prefer

Answer 8

i think that u got it..

Answer 9

keep right side, solve left side?

Answer 10

?

Answer 11

you can do that also

Answer 12

\[\frac{1+\tan \theta}{\sin \theta+\cos \theta }=\frac{\cos \theta (1 +\tan \theta)}{\cos \theta (\sin \theta +\cos \theta)}\]\[\frac{\sin \theta + \cos \theta}{\cos \theta(\sin \theta +\cos \theta)}=1/\cos \theta =\sec \theta =RHS\]

Answer 13

I don't get

Answer 14

\[\frac{\cos \theta+\sin \theta}{\frac{\cos \theta}{\sin \theta+\cos \theta}}\]I get from here

Answer 15

\[=\frac{(\cos \theta+\sin \theta)^2}{\cos \theta}\]

Answer 16

what are you doing????

Answer 17

start from any one of the sides..either RHS or LHS

Answer 18

I do
1+tan

Answer 19

take 1+tan/(sin + cos)

Answer 20

multiply num and den with cos and see my previous post

Answer 21

now got it?