cos-cot-sin-tan-csc-sec-

Question Number 178181 by mathlove last updated on 13/Oct/22

\frac{c o s α c o t α - s i n α t a n α}{c s c α s e c α} = ?

Commented by Adedayo2000 last updated on 13/Oct/22

R e c a l l c o t α = \frac{1}{t a n α} = \frac{c o s α}{s i n α}, c s c α = \frac{1}{s i n α}, s e c α = \frac{1}{c o s α}

=> \frac{(c o s α) (\frac{c o s α}{s i n α}) - (s i n α) (\frac{s i n α}{c o s α})}{(\frac{1}{s i n α}) (\frac{1}{c o s α})}

=> \frac{\frac{{c o s}^{2} α}{s i n α} - \frac{{s i n}^{2} α}{c o s α}}{\frac{1}{s i n α c o s α}} => \frac{\frac{{c o s}^{3} α - {s i n}^{3} α}{s i n α c o s α}}{\frac{1}{s i n α c o s α}}

=> {c o s}^{3} α - {s i n}^{3} α

A δ ξ δ α Υ o Δ δ ξ β α γ o \dots .

Answered by Ar Brandon last updated on 13/Oct/22

\frac{\cos α \cot α - \sin α \tan α}{\csc α \sec α} = \cos^{3} α - \sin^{3} α

Answered by itzikhaim last updated on 13/Oct/22

= \frac{c o s α \frac{c o s α}{s i n α} - s i n α \frac{s i n α}{c o s α}}{c s c α s e c α}

= \frac{\frac{{c o s}^{2} α}{s i n α} - \frac{{s i n}^{2} α}{c o s α}}{(\frac{1}{c o s α}) (\frac{1}{s i n α})} = \frac{\frac{{c o s}^{2} α - {s i n}^{2} α}{s i n α c o s α}}{\frac{1}{c o s α s i n α}}

= {c o s}^{2} α - {s i n}^{2} α

= c o s (2 α)

Commented by Ar Brandon last updated on 13/Oct/22

\frac{\cos^{2} α}{\sin α} - \frac{\sin^{2} α}{\cos α} = \frac{\cos^{3} α - \sin^{3} α}{\sin α \cos α}