5sec-4-tan-3cosec-3cot-5-tan-4-sec-

Question Number 161664 by cortano last updated on 21/Dec/21

5 \sec α - 4 \tan α = 3 cosec α

\frac{3 \cot α}{5 \tan α - 4 \sec α} = ?

Answered by som(math1967) last updated on 21/Dec/21

{(5 s e c α - 4 t a n α)}^{2} = 9 {c o s e c}^{2} α

o r 25 {s e c}^{2} α + 16 {t a n}^{2} α - 40 s e c α t a n α

= 9 + 9 {c o t}^{2} α

o r 25 + 25 {t a n}^{2} α + 16 {s e c}^{2} α - 16 - 40 t a n α s e c α

= 9 + 9 {c o t}^{2} α

o r {(5 t a n α - 4 s e c α)}^{2} = {(3 c o t α)}^{2}

o r (5 t a n α - 4 s e c α) = \pm 3 c o t α

∴ \frac{3 c o t α}{5 t a n α - 4 s e c α} = \pm 1