Question

If (1 + tan 5/(1 + tan 10°)(1 + tan 159........ (1 + tan 45º) (1 + tan 225º)=2 raise to k
then 'K' equals​

Answer 1

Correct Question

If ( 1 + tan 5° ) ( 1 + tan 10° ) ( 1 + tan 15° ) . . . . . . . . . . . ( 1 + tan 45° ) = 2^k . Find the value of k ?

ANSWER

Given :

If ( 1 + tan 5° ) ( 1 + tan 10° ) ( 1 + tan 15° ) . . . . . . . . . . ( 1 + tan 45° ) = 2^k

Required to find :

• Value of k ?

Formula used :

tan ( A + B ) = tan A + tan B/1 - tan A tan B

Solution :

If ( 1 + tan 5° ) ( 1 + tan 10° ) ( 1 + tan 15° ) . . . . . . . . . . ( 1 + tan 45° ) = 2^k

We need to find the value of k ?

So,

Consider the LHS part

( 1 + tan 5° ) , ( 1 + tan 10° ) , ( 1 + tan 15° ) , . . . . . . . . . . . . . . , ( 1 + tan 45° )

This can be written as ;

[ ( 1 + tan 5° ) ( 1 + tan 40° ) ] [ ( 1 + tan 10° ) ( 1 + tan 35° ) ] [ ( 1 + tan 15° ) ( 1 + tan 30° ) ] [ ( 1 + tan 20° ) ( 1 + tan 25° ) ] ( 1 + tan 45° )

Now,

We know that ;

tan ( A + B ) = tan A + tan B/1 - tan A tan B

This implies ;

tan A + tan B = tan ( A + B ) x 1 - tan A tan B

Here, let's consider the first part of the LHS as A .

➜ A = ( 1 + tan 5° ) ( 1 + tan 40° )

Solving further !

➜ 1 ( 1 + tan 40° ) + tan 5° ( 1 + tan 40° )

➜ 1 + tan 40° + tan 5° + tan 5° tan 40°

Here,

tan 40° + tan 5° is in the form of tan A + tan B . So, from the above formula we can write this as ;

➜ 1 + tan ( 40° + 5° ) x 1 - tan 40° tan 5° + tan 5° tan 40°

➜ 1 + tan 45° x 1 - tan 40° tan 5° + tan 5° tan 40°

Since,

• tan 45° = 1

➜ 1 + 1 x 1 - tan 40° tan 5° + tan 5° tan 40°

- tan 40° tan 5° , tan 40° tan 5° get's cancelled

➜ 1 + 1

➜ 2

Similarly,

Consider the 2nd part in the LHS part as B

➜ B = ( 1 + tan 10° ) ( 1 + tan 35° )

➜ 1 ( 1 + tan 35° ) + tan 10° ( 1 + tan 35° )

➜ 1 + tan 35° + tan 10° + tan 35° tan 10°

➜ 1 + tan ( 35° + 10° ) x 1 - tan 35° tan 10° + tan 10° tan 35°

➜ 1 + tan 45°

➜ 1 + 1

➜ 2

Similarly,

Consider the 3rd part in the LHS part as C

➜ C = ( 1 + tan 15° ) ( 1 + tan 30° )

➜ 1 ( 1 + tan 30° ) + tan 15° ( 1 + tan 30° )

➜ 1 + tan 30° + tan 15° + tan 30° tan 15°

➜ 1 + tan ( 30° + 15° ) x - 1 - tan 30° tan 15° + tan 30° tan 15°

➜ 1 + tan 45°

➜ 1 + 1

➜ 2

Similarly,

Consider the 4th part in the LHS part as D

➜ D = ( 1 + tan 20° ) ( 1 + tan 25° )

➜ 1 ( 1 + tan 25° ) + tan 20° ( 1 + tan 25° )

➜ 1 + tan 25° + tan 20° + tan 25° tan 20°

➜ 1 + tan ( 25° + 20° ) x 1 - tan 25° tan 20° + tan 25° tan 20°

➜ 1 + tan 45°

➜ 1 + 1

➜ 2

Similarly,

Consider the 5th part in the LHS part as E

➜ E = ( 1 + tan 45° )

➜ 1 + 1

➜ 2

Now,

Let's integrate the whole !

➜ [ ( 1 + tan 5° ) ( 1 + tan 40° ) ] [ ( 1 + tan 10° ) ( 1 + tan 35° ) ] [ ( 1 + tan 15° ) ( 1 + tan 30° ) ] [ ( 1 + tan 20° ) ( 1 + tan 25° ) ] ( 1 + tan 45° )

➜ A x B x C x D X E

➜ 2 x 2 x 2 x 2 x 2

➜ 2^5

Equate the LHS and RHS

➜ 2^5 = 2^k

Since bases are equal powers are also equal

➜ k = 5

Therefore,

• Value of k = 5