Question -
Answer -
(i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x
Let us consider LHS:
tan 8x – tan 6x – tan 2x
tan 8x = tan(6x + 2x)
We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)
So,
tan 8x = (tan 6x + tan 2x) / (1 – tan 6x tan 2x)
By 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
Upon rearranging we get,
tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x
= RHS
∴ LHS = RHS
Hence proved.
(ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1
We know,
π/12 = 15° and π/6 = 30°
So, we have 15° + 30° = 45°
Tan (15° + 30°) = tan 45°
We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)
So,
(tan 15o + tan 30o) / (1 –tan 15o tan 30o) = 1
tan 15° + tan 30° = 1 – tan 15° tan 30°
Upon rearranging we get,
tan15° + tan30° + tan15° tan30° = 1
Hence proved.
(iii) tan 36o + tan 9o + tan36o tan 9o = 1
We know 36° + 9° = 45°
So we have,
tan (36° + 9°) = tan 45°
We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)
So,
(tan 36o + tan 9o) / (1 –tan 36o tan 9o) = 1
tan 36° + tan 9° = 1 – tan 36° tan 9°
Upon rearranging we get,
tan 36° + tan 9° + tan 36° tan 9° = 1
Hence proved.
(iv) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x
Let us consider LHS:
tan 13x – tan 9x – tan 4x
tan 13x = tan (9x + 4x)
We know that, tan (A + B) = (tan A + tan B) / (1 – tanA tan B)
So,
tan 13x = (tan 9x + tan 4x) / (1 – tan 9x tan 4x)
By 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
Upon rearranging we get,
tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x
= RHS
∴ LHS = RHS
Hence proved.