title

D.

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A is easy,

Work on the LHS to prove it is equal to the RHS

Multiply it by    $$\\(1+cos\theta)/(1+cos\theta)\\$$

$$\\$you will also need to recognise that $\\sin^2\theta+cos^2\theta=1$$

I will leave you to work out the details  :))

I'm sorry. But, what is LHS and RHS? Help me out, please? :(

I am sorry

LHS. Left hand side

RHS.  Right hand side

work on the LHS until it equals the RHS.  :)

Start by doing the multiplication that I suggested :)

A.  [sin^2 Θ ] / [1 - cos Θ ]      

[1 - cos^2 Θ ] /  [ 1 - cos Θ ] =

[(1 + cos Θ )(1 - cos Θ ) ]  / [ 1 - cos Θ ] =

1 + cos Θ

B.   Working on the left hand side

[ cot Θ cos Θ ] / [ cot Θ  + cos Θ ]  =

[(cos Θ / sin Θ ) cos Θ ] / [ (cos Θ /  sin Θ ) + cos Θ ] =

[ (cos^2 Θ  / sin Θ ) ] /  [ (cos Θ  + sin Θ  cos Θ ) / sin Θ  ]

[cos^2 Θ ] / [ cos Θ ( 1 + sin Θ ) ]

[ (1 -  sin^2 Θ ) ] / [cos Θ  (1 + sin Θ ) ]

[ (1 - sin Θ ) ( 1 + sin Θ ) ] /   [cos Θ  ( 1 + sin Θ ) ]

[ 1 - sin Θ ]   /  cos Θ           multiply top and bottom by cot Θ

[cot Θ  -  cos Θ ] / [cot Θ cos Θ  ]    and that equals the right hand side

C.....this one is similar to A.......

cos^2 Θ / [ 1 + sin Θ  ]

[ 1 - sin^2 Θ ] / [ 1 + sin Θ ]

[(1 - sin Θ ) ( 1 + sin Θ ) ] / [ 1 + sin Θ ]

1 - sin Θ     = the right hand side