Question

tan30+tan 45+tan 60+tan 15

Answer 1

hope it helps u dear

Answer 2

Hey friend !
Here is your answer!

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tan 30° + tan 45° + tan 60° + tan 15°

=> 1/√3 + 1 +√3 + tan 15°

=> Now solve the tan 15°

Let tan(15°) = tan(45°-30°)

We know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)

⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°)

= {1- (1/√3)} / {1+(1/√3)}

∴ tan15° = (√3 - 1) / (√3 + 1)

Then,

1/√3 + 1 +√3 + tan 15°
=>
1/√3 + 1 +√3 +(√3 - 1) / (√3 + 1)

Now take LCM √3(√3+1)

[√3+1 +√3(√3+1) + √3×√3 ×(√3+1)+√3(√3-1) ]. / √3(√3+1)

[√3+1 +3+√3 +3√3+3 +3-√3] /3+√3

Now cancelled +√3 &-√3
=>
[√3+3√3 +10] / 3+√3

[4√3 +10] / 3+ √3

Put the value of √3= 1.73
[4×1.73 +10] / 3+ 1.73

[6.92 +10] /4.73

=> 16.92 / 4.73

=> 3.43 Answer
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Hope it's helpful!