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Grade 11,

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Prove that tan75 + cot75 = 4

Prove that tan75 + cot75 = 4

Grade:12

7 Answers

Pulkit Singhal
25 Points
12 years ago

tan75o = sqrt(3) + 1 / sqrt(3) - 1 

cot 75 = sqrt(3) - 1 / sqrt(3) + 1

on adding both of these we get

tan75 + cot75 = (sqrt(3) + 1)^2+(sqrt(3) - 1)^2  / (sqrt(3) - 1)(sqrt(3) + 1)

on solving you get answer 4

Please approve it if you like

Menka Malguri
39 Points
12 years ago

tan75°=tan(30°+45°)

=(tan30°+tan45°)/(1-tan30°tan45°)

={(1/√3)+1}/{1-(1/√3)(1)}

=(√3+1)/(√3-1)

So,cot75°=1/tan75°

=(√3-1)/(√3+1)

Then,tan75°+cot75°={(√3+1)2+(√3-1)2}/{(√3)2-(1)2}

=(3+1+2√3+3+1-2√3)/(3-1)

=4.

ankitesh gupta
63 Points
11 years ago

 

 

PLZZZZZZZZZZZZZZZZZ APPROVE THE METHOD IF IT HELPED YOU BY CLICKING THE YES BUTTON........................................Smile

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Niharika
13 Points
6 years ago
The value of tan 75°=2-root3It`s very easy and simple just the thing is you have to remember the formula. I will say you in a easy and simple methodThe value of cot 75°=2+root3Tan75°+cot75°2-root3+2+root3=4I hope you got the answer
Sivaranjani
15 Points
5 years ago
Sin 75 degree- sin 15 degree = cos 105 degree + cos 15 degree .          How to solve in trignotrigon
 For 11 standard 
Vaisakh
15 Points
5 years ago
tan75=2+√3.
Cot75=2-√3.
So tan75+cot75=2+√3+2-√3=4.
TThis isa very easy method for doing this so i advixe u do by this method...Thank u
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem below.
 
tan75°= tan(30°+45°)
=(tan30°+tan45°)/(1-tan30°tan45°)
={(1/√3)+1}/{1-(1/√3)(1)}
=(√3+1)/(√3-1)
 
So,cot75°=1/tan75°
=(√3-1)/(√3+1)
Then, tan75°+cot75°={(√3+1)2+(√3-1)2}/{(√3)2-(1)2}
=(3+1+2√3+3+1-2√3)/(3-1)
=4.
 
Thanks and regards,
Kushagra

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