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prove tan15° + cot15° = 4

Let α = 15°, then we have the LHS =

tan(α) + cot(α) =

sin(α)/cos(α) + cos(α)/sin(α) =

{sin^2(α) + cos^2(α)} / (sin(α).cos(α)} =

1 / (sin(α).cos(α)} =

2 / (2.sin(α).cos(α)} =

2 / (sin(2α)} =

2 / (1/2} =     (since α = 15°, and sin(30) = ½)

4 = RHS

by Level 11 User (81.5k points)

tan15=tan(45-30)=tan45-tan30/1+tan45.tan30=1-(1/√3)/1+1/√3=√3-1/√3+1

         =2-√3    (after rationalisation)

similarly

cot15=2+√3

 

so

tan45+cot45=2+√3+2-√3

                    =4

by Level 1 User (260 points)

Let A =tan15° + Cot15° =tan15° + 1/tan15°

A = (tan²15° + 1)/tan15° ………….. (1)

Now, we know that: tan30°=tan(2.15°)

i.e. tan (2.15°) = 2tan15°/(1 - tan²15°)

i.e. 2tan15°/(1 - tan²15°) = 1/√3

(Note: tan30° = 1/√3)

i.e. tan15° = 1/2√3(1 - tan²15°)

i.e. tan15° = √3/6(1 - tan²15°) ……… (2)

Substitute (2) in (1), we have:

A =(1 + tan²15°)/{√3/6(1 - tan²15°)} …(3)

A= 6(1 + tan²15°)/√3(1 - tan²15°)

Expand the following terms in (3) below:

1 + tan²15° = 1 + Sin²15°/Cos²15° ……. (i)

= (Sin² + Cos²x)/Cos²x

1 - tan²15° = 1 - Sin²15°/Cos²15° …….. (ii)

= (Cos²15° - Sin²15°)/Cos²15°

Substitute (i) & (ii) in (4), we have:

A=(6/√3){Sin²15°+Cos²15°}/{(Cos²15°-Sin²15°)} x Cos²15°/Cos²15°

Recall that: Sin²15° + Cos²15° = 1……(4)

i.e. A = {(6/√3) x 1}/(Cos²15° - Sin²15°) …(5)

Cos30°=Cos(2.15°)=Cos²15°-Sin²15°

And, Cos30° = √3/2

i.e. Cos2.15°=Cos²15°- Sin²15°=√3/2 …(6)

Substitute (6) in (5), we have:

A = (6/√3)/(√3/2)

A = (6/√3) x (2/√3) = (6 x 2)/(√3 x √3)

A = 12/3 = 4

Hence: tan15° + Cot15° = 4

baccalaureateclass

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