Question

Prove the following:

i. sec⁶ x – tan⁶ x = 1 + 3 sec² x × tan² x

ii. tanθ/(secθ + 1) = (secθ - 1)/tanθ

Answer 1

L.H.S. = sec⁶x – tan⁶x 

= (sec²x)³ – tan⁶x 

= (1 + tan²x)³ – tan⁶x …[∵ 1 + tan²θ = sec²θ] 

= 1 + 3tan² x + 3(tan²x)² + (tan²x)³ – tan⁶x …[∵ (a + b)³ = a³ + 3a² b + 3ab² + b³]

= 1 + 3 tan²x (1 + tan²x) + tan⁶x – tan⁶ x 

= 1 + 3 tan²x sec²x …[∵ 1 + tan²θ = sec²θ] 

= R.H.S. 

∴ sec³x – tan⁶x = 1 + 3sec²x.tan²x

ii. L.H.S. = tanθ/(secθ + 1)