(i) Let us consider the LHS

tan 8x – tan 6x – tan 2x

tan 8x = tan(6x + 2x)

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

tan 8x = (tan 6x + tan 2x)/(1 – tan 6x tan 2x)

On cross-multiplying we get,

tan 8x(1 – tan 6x tan 2x) = tan 6x + tan 2x

tan 8x – tan 8x tan 6x tan2x = tan 6x + tan 2x

Now, upon rearranging we get,

tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x

= RHS

∴ LHS = RHS

Thus proved.

(ii) As we know,

π/12 = 15° and π/6 = 30°

Therefore, we have 15° + 30° = 45°

tan(15° + 30°) = tan 45°

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 15° + tan 30°)/(1 – tan 15° tan 30°) = 1

tan 15° + tan 30° = 1 – tan 15° tan 30°

Now, upon rearranging we get,

tan15° + tan30° + tan15° tan30° = 1

Thus proved.

(iii) As we know 36° + 9° = 45°

Therefore we have,

tan (36° + 9°) = tan 45°

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

(tan 36° + tan 9°)/(1 – tan 36° tan 9°) = 1

tan 36° + tan 9° = 1 – tan 36° tan 9°

Now, upon rearranging we get,

tan 36° + tan 9° + tan 36° tan 9° = 1

Thus proved.

(iv) Let us consider the LHS

tan 13x – tan 9x – tan 4x

tan 13x = tan (9x + 4x)

As we know that, tan (A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore,

tan 13x = (tan 9x + tan 4x)/(1 – tan 9x tan 4x)

On cross-multiplying we get,

tan 13x (1 – tan 9x tan 4x) = tan 9x + tan 4x

tan 13x – tan 13x tan 9x tan 4x = tan 9x + tan 4x

Now, upon rearranging we get,

tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x

= RHS

∴ LHS = RHS

Thus proved.