Question

Value of (root3+tan1)(root3+tan2).............(root3+tan29)​

Answer 1

Answer:

...................

Answer 2

Given: The term (√3+tan 1°)(√3+tan 2°).............(√3+tan 29°)​

To find: The value of the given series.

Solution:

• Now we have given :

                (√3+tan 1°)(√3+tan 2°).............(√3+tan 29°)​

• Now consider a term:

                √3 + tan(30 - x)

• Expanding it, we get:

                √3 + tan(30 - x) = √3 + { (1/√3) - tanx / 1 + (tanx/√3) }

                √3 + tan(30 - x) = √3 + { (1 - √3tanx /√3 + tanx }

                √3 + tan(30 - x) = 4 / (√3 + tanx)

                {√3 + tan(30 - x)} (√3 + tanx) = 4

• Putting x = 1, we get:

                {√3 + tan(29)} (√3 + tan1) = 4    .......(i)

• Putting x = 2, we get:

                {√3 + tan(28)} (√3 + tan2) = 4      ...........(ii)

• Similarly by putting x = 1-14, we will get all the terms which are present in the given series, except for (√3+tan 15°).

• So putting 15 now, we get:

                {√3 + tan(15)} (√3 + tan 15) = 4

• Here we can see that both the brackets are same, so:

                {√3 + tan(15)}² = 4

                {√3 + tan(15)} = 2    (positive because tan 15 > 0) ........(iii)

• Multiplying (i), (ii) , all terms till 14 and (iii), we get:

                (√3+tan 1°)(√3+tan 2°).............(√3+tan 29°)​ = 4^14 x 2

                (√3+tan 1°)(√3+tan 2°).............(√3+tan 29°)​ = 2^29

Answer:

              So the value of the given series is 2^29.