Can You Simplify a Trigonometric Expression? | Two Methods

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[GauravJha-mugv]

Divide numerator and denominator of second term by cos x. It would be much simpler.

[theelk]

I read the thumbnail, solved it very quickly, then wondered how you managed to get two different methods

[mbmillermo]

The first method didn't occur to me, thankfully!

[pluisjenijn]

Why not divide the second part with cos x / cos x. Then you have the same denominator 1+tan x. You can then add the numerators and immediately you see the answer is 1

[josephshaff]

Sweet ! 😅 I was so close ! I figured it for 1. If it was multiple choice I would've put 1. Good one, Good one !

[wonghonkongjames]

Good evening, Sir,
Is the answer 1, I suppose there may be other solution to choose

[alextang]

Change tan(x) to sin(x)/cos(x), problem is solved. 😋😋😋😋😋😋

[arnavmeena]

Just write tanx as sinx/cosx and take lcm. Bruh.

[sumithpeiris]

can do without pen or paper just by dividing numerator and denominator of 2nd fraction by cosx

[SidneiMV]

divide both numerator and denominator of the second fraction by cosx and we get tanx / (1 + tanx) and the problem is solved.

[sumam]

Just write tan x in terms of cos x and send sin x and take LCM and ans. is
..1

[toveirenestrand]

1/(1 + tanx) + sinx/(cosx + sinx) = 1/[(cosx + sinx)/cosx] + sinx/(cosx + sinx) = cosx/(cosx + sinx) + sinx/(cosx + sinx) = (cosx + sinx)/(cosx + sinx) = 1.

[kallolsinha]

We know tanx=sinx/cosx put the value of tanx putting this thing and take some LCM and solve it by three steps. It's very easy not complicated.

[barberickarc]

1+tg(x) = cos(x) / cos (x) + sin (x) /cos (x)
Adding them together you get [cos (x) + sin (x)] / cos (x).
To find the reciprocal just flip,
1 / [ 1 + tg (x)] = cos (x) / [ cos (x) + sin (x)]
Plug in to the original equation and get
[cos (x) + sin (x)] / [ cos (x) + sin (x)] = 1

[biscuit_]

Or write 1/(1+tanx) as cosx/(cosx + sinx) [by writing tanx as sinx/cosx] and now we have
cosx/(cosx+sinx) + sinx/(cosx + sinx)=1

[nidhisrivastava]

Expressing
Tanx in the terms of sinx and cosx

Tanx = Sinx/cosx

How?
LET ABC BE A RIGHT ANGLED AT A AND ANGLE B BE x
Sinx= PERPENDICULAR/ HYPOTENUSE

Cosx = BASE/ HYPOTENUSE

TAKING RATIO
Sinx/cosx= PERPENDICULAR/BASE = Tanx

[AnibalRiba]

If x = 3*pi/4 + k*pi this is not going to work, as it will be infinite.