prove this TAN + COT = SEC + CSC?

Answer 1

By themselves, they are meaningless (because they are functions). They need to be applied to some variable, x.

You might have made an error when copying it down.
You want to prove the identity

tan(x) + cot(x) = sec(x) csc(x)

Work on the more complex side first. That will be the left hand side.

LHS = tan(x) + cot(x)

Convert everything to sines and cosines.

LHS = [sin(x)/cos(x)] + [cos(x)/sin(x)]

Merge into one fraction by finding the greatest common denominator.

LHS = [sin^2(x) + cos^2(x)]/[sin(x)cos(x)]

The numerator is equal to 1.

LHS = 1/[sin(x)cos(x)]

Split into a product of fractions.

LHS = (1/sin(x)) (1/cos(x))
LHS = csc(x)sec(x)

Multiplication is commutative.

LHS = sec(x)csc(x) = RHS

Answer 2

TAN + COT = sin / cos + cos / sin

= ( sn^2 + cos^2) /sin cos
= 1 / sin cos
= 1/ sin x 1/cos
= sec cosec
that is the right answer

Answer 3

tan + cot = sin/cos + cos/sin
= sin2/cossin + cos2/cossin
= sin2 + cos2/cossin = 1/cossin
= 1/cos 1/sin = sec + csc

Answer 4

LHS
sin x/cos x + cos x/sin x
(sin² x + cos² x) / (cos x.sin x)
(1/cos x).(1 / sin x)
sec x . cosec x
RHS
secx .cosec x ? (error in question?)
LHS = RHS

Answer 5

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