Question

Prove That
(cosecA-sinA)(secA-cosA)(tanA+cotA) = 2

Answer 1

@k_lynn

Answer 2

Sorry, I"m not very good at that kind of math.

Answer 3

\((\csc A- \sin A)(\sec A- \cos A)(\tan A+ \cot A) = 2\)

\((\dfrac{1}{\sin A} - \sin A)(\dfrac{1}{\cos A}- \cos A)(\dfrac{\sin A}{\cos A}+ \dfrac{\cos A}{\sin A}) \)

\((\dfrac{1}{\sin A} - \sin A \times \dfrac{\sin A}{\sin A})(\dfrac{1}{\cos A}- \cos A \times \dfrac{\cos A}{\cos A})(\dfrac{\sin A}{\cos A} \times \dfrac{\sin A}{\sin A}+ \dfrac{\cos A}{\sin A}\times \dfrac{\cos A}{\cos A}) \)

Can you continue?

Answer 4

Sorry, the last line above was cut off. Here it is:

\((\dfrac{1}{\sin A} - \sin A \times \dfrac{\sin A}{\sin A})(\dfrac{1}{\cos A}- \cos A \times \dfrac{\cos A}{\cos A}) \)

\( \times (\dfrac{\sin A}{\cos A} \times \dfrac{\sin A}{\sin A}+ \dfrac{\cos A}{\sin A}\times \dfrac{\cos A}{\cos A}) \)

Answer 5

In each set of parentheses, add or subtract the fractions using the common denominators written. Then use trig identity substitutions.

Answer 6

??????????

Answer 7

What???

Answer 8

@mathmate

Answer 9

@rvc

Answer 10

@satellite73

Answer 11

@jigglypuff314

Answer 12

\((\dfrac{1}{\sin A} - \sin A \times \dfrac{\sin A}{\sin A})(\dfrac{1}{\cos A}- \cos A \times \dfrac{\cos A}{\cos A})\)

\(\times (\dfrac{\sin A}{\cos A} \times \dfrac{\sin A}{\sin A}+ \dfrac{\cos A}{\sin A}\times \dfrac{\cos A}{\cos A})\)


\((\dfrac{1}{\sin A} - \dfrac{\sin^2 A}{\sin A})(\dfrac{1}{\cos A}- \dfrac{\cos^2 A}{\cos A}) (\dfrac{\sin^2 A}{\cos A \sin A} + \dfrac{\cos^2 A}{\cos A\sin A})\)

\((\dfrac{1 -\sin^2 A}{\sin A})(\dfrac{1 - \cos^2 A}{\cos A}) (\dfrac{\sin^2 A + \cos^2 A}{\cos A\sin A})\)

\((\dfrac{\cos^2 A}{\sin A})(\dfrac{\sin^2 A}{\cos A}) (\dfrac{1}{\cos A\sin A})\)

\(1\)

I get 1 on the left side, not 2.