I just had a test, and one question I tried heaps of times and I still couldn't answer. It was "Prove secèsinè + cosecè cosè = 2cosecè "

Can someone please tell me if proving this is possible at all? (But don't tell me how to solve it)

sin x/cos x + cos x/sin x =

(sin^2 x + cos^2 x)/(sin x cos x )=
1/(sin x cos x)
I do not see how that comes out to 2 cosec x

Usually my first step in proving any trig identity is taking an arbitrary angle to see if it makes it true, say 45 degrees.

Picking up from Damon's first line, your equation is
tanx + cotx = 2/sinx

for x = 45 degrees
LS = 1+1 = 2
RS = 2/.707 which is not LS

so the equation is not an identity.

Yes, it is possible to prove the given equation: sec(θ) + cosec(θ) = 2cosec(θ). This can be done using trigonometric identities and algebraic manipulations.

To prove this equation, we will start with the left-hand side (LHS) and manipulate it using trigonometric identities to obtain the right-hand side (RHS).

First, let's write the LHS in terms of sine and cosine:

sec(θ) + cosec(θ) = 1/cos(θ) + 1/sin(θ)

Now, let's find a common denominator for the two terms on the LHS:

(1/cos(θ)) * (sin(θ)/sin(θ)) + (1/sin(θ)) * (cos(θ)/cos(θ))
= sin(θ)/(sin(θ) * cos(θ)) + cos(θ)/(sin(θ) * cos(θ))

Now, let's combine the two terms on the LHS by finding a common denominator:

(sin(θ) + cos(θ))/(sin(θ) * cos(θ))

Finally, let's simplify the numerator by using the trigonometric identity sin(θ) + cos(θ) = √2 * sin(θ + π/4):

= (√2 * sin(θ + π/4))/(sin(θ) * cos(θ))

This expression is not equal to 2cosec(θ) in general. Hence, the given equation sec(θ) + cosec(θ) = 2cosec(θ) is not true for all values of θ.

Therefore, the equation sec(θ) + cosec(θ) = 2cosec(θ) cannot be proven true.