Question
Solve the trigonometric equations.
1.) SIN²x(CSC²x-1)
2.) COT x SEC X
3.) COS²[(π/2)-x]/COS X
(Note: These mathematical problems are somewhat tricky, but useful for students as of learning how to their fundamental identities. Let's enjoy solving these problems!)
1.) SIN²x(CSC²x-1)
2.) COT x SEC X
3.) COS²[(π/2)-x]/COS X
(Note: These mathematical problems are somewhat tricky, but useful for students as of learning how to their fundamental identities. Let's enjoy solving these problems!)
Answers
You have no equations to solve, you probably meant "simplify the expressions"
1. sin^2 x (1/sin^2 x - 1)
= 1 - sin^2 x
= cos^2 x
2. (cosx/sinx) (1/cosx)
= 1/sinx
= cscx
3. recall the property of complementary trig ratios, that is...
sin(π/2 - x) = cosx and cos(π/2 - x) = sinx
sec(π/2 - x) = cscx and .....
tan(π/2 - x) = cotx
notice the names:
sine --co-sine or cosine
tangent -- co-tangent or cotangent
secant --- co-secant or cosecant
e.g. sin (90° - 20°) = sin 70° = cos20°
so your question:
COS²[(π/2)-x]/COS X
= sin^2 x / cosx
= (sinx)(sinx/cosx)
= sinx tanx
1. sin^2 x (1/sin^2 x - 1)
= 1 - sin^2 x
= cos^2 x
2. (cosx/sinx) (1/cosx)
= 1/sinx
= cscx
3. recall the property of complementary trig ratios, that is...
sin(π/2 - x) = cosx and cos(π/2 - x) = sinx
sec(π/2 - x) = cscx and .....
tan(π/2 - x) = cotx
notice the names:
sine --co-sine or cosine
tangent -- co-tangent or cotangent
secant --- co-secant or cosecant
e.g. sin (90° - 20°) = sin 70° = cos20°
so your question:
COS²[(π/2)-x]/COS X
= sin^2 x / cosx
= (sinx)(sinx/cosx)
= sinx tanx