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
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 answer