Question

Prove ((cscx-cotx)(cscx+cotx))/secx=cosx

Trigonometric equation and identitie
ANSWERED
asked 2021-01-28
Prove \(\displaystyle\frac{{{\left({\csc{{x}}}-{\cot{{x}}}\right)}{\left({\csc{{x}}}+{\cot{{x}}}\right)}}}{{\sec{{x}}}}={\cos{{x}}}\)

Answers (1)

2021-01-29

\(\displaystyle{L}{H}{S}=\frac{{{{\csc}^{{2}}{x}}-{\cot}^{{2}}}}{{\sec{{x}}}}{\left({d}\Leftrightarrow\text{erence of two}\right)}\)
=\(1/ \sec x ( \text{formula }1 + \cot^2 x = \csc^2 x , \text{thus }\csc^x - \cot^2 x = 1)\)
\(\displaystyle={\cos{{x}}}={R}{H}{S}\)

0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...