Math, asked by laservictor2018, 1 year ago

prove cot70°+4cos70°=√3

Answers

Answered by YASH3100

Here is ur ans,
LHS => cot70+4cos70 = tan20+4sin20 [using cot(x) = tan(90-x) and cos(x) = sin(90-x)]
= (sin20+4sin20cos20)/cos20
= (sin20+2sin40)/cos20 [using 2sin(x)cos(x) = sin(2x)]
= (sin20+sin40+sin40)/cos20
= (2sin30cos10+sin40)/cos20 [using sin(C)+sin(D) = 2sin((C+D)/2)cos(C-D)/2)]
= (cos10+cos50)/cos20 [using sin30 = 1/2 ]
= (2cos30cos20)/cos20 [using cos(C)+cos(D) = 2cos((C+D)/2)cos((C-D)/2)]
= 2cos30
= √3 = RHS (answer)

Hope it helps you.
Thank you.

laservictor2018: this is brainliest answer

YASH3100: Then pls do add it also na as brainliest

laservictor2018: but how.... no one else has answered it yet

YASH3100: Haan that's the prob na and I don't think so that anyone else would answer it.

YASH3100: Let it be for me it's more Imp that you got ur doubt cleared.

laservictor2018: thanx a lot for the answer

laservictor2018: btw

YASH3100: Are most welcome brother

laservictor2018: followed u

YASH3100: Thank you so much bro.

Answered by ultimatebeast

Answer: $\sqrt{3$

Step-by-step explanation:LHS => cot70+4cos70 = tan20+4sin20 [using cot(x) = tan(90-x) and cos(x) = sin(90-x)]

= (sin20+4sin20cos20)/cos20

= (sin20+2sin40)/cos20 [using 2sin(x)cos(x) = sin(2x)]

= (sin20+sin40+sin40)/cos20

= (2sin30cos10+sin40)/cos20 [using sin(C)+sin(D) = 2sin((C+D)/2)cos(C-D)/2)]

= (cos10+cos50)/cos20 [using sin30 = 1/2 ]

= (2cos30cos20)/cos20 [using cos(C)+cos(D) = 2cos((C+D)/2)cos((C-D)/2)]

= 2cos30

= √3 = RHS (answer)

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