How do I show that this expression is equivalent?

Question

1 + sec w / tan w + sin w = csc w

I made an attempt to solve this problem, but I can't solve it. Please help me!

Answer 1

LHS =
(1+ sec w)/(tan w + sin w)
= (1+ sec w)/(sin w/cosw + sin w)
= (1+ sec w )/ (sin w (1/cosw + 1))
= (1+ sec w )/ (sin w(sec w + 1))

= 1/ sin w
= csc w

Answer 2

In MOST identity-solving problems, the first two steps are
1) Choose the most complex side,
2) Change everything to sines and cosines.

In this case, we choose the left hand side (LHS), because it is obviously more complex.

LHS = [1 + sec(w)]/[tan(w) + sin(w)]

Change everything to sines and cosines, keeping in mind the definition of secant (1 over cos) and tan (sin over cos).

LHS = [1 + 1/cos(w)]/[ sin(w)/cos(w) + sin(w) ]

Now, multiply top and bottom by cos(w), to eliminate all fractions.

LHS = [cos(w) + 1] / [sin(w) + cos(w)sin(w)]

In the denominator, factor out the common term, sin(w), toget

LHS = [cos(w) + 1]/[ sin(w) [1 + cos(w)] ]

Note that addition is commutative (a + b = b + a), thus we can rearrange 1 + cos(w)

LHS = [cos(w) + 1]/[ sin(w) [cos(w) + 1] ]

And look! The numerator and denominator have something in common! We can effectively cancel them, leaving us with

LHS = 1/sin(w) = csc(w) = RHS

Therefore, this expression is equivalent.

Answer 3

You should use brackets for more clarity in the problem.
The problem should like this, I suppose .....
(1 + sec w) / (tan w + sin w) = Cosec w.
Ok ..... let us take LHS and convert all into Sin and Cos ....
(1 + 1/cos w) / (sin w / cos w + sin w)
= [(Cos w + 1)/ Cos w] / {(Sin w + Sin w Cos w)/Cos w}
= (Cos w +1)/ Cos w x Cos w / [Sin w (1+Cos w)] .... then cancel the similar terms in the numerator and denominator .....
= 1/ Sin w = Cosec w