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One equality Common Denominator Form

Problem:

I'm trying to prove the following:

using the following identity:

I have tried various methods, but I am baffled!

Incidentally, I would be very grateful if you could also show me how to prove the second formula.

Solution:

We can solve following identity without using the second one.

First we have to expand the left-hand side term by using the formula

nCr = (n)/[(r)  (n-r) ]

Applying in left-hand side we will get
LHS = n-1Ck-m-1*n-1Cm-1 + n-1Ck-m-2*n-1Cm +n-1Ck-m-3*n-1Cm+1 +.......

= (n-1)/(k-m-1)(n-k+m) * (n-1)/(m-1)(n-m) + (n-1)/(n-k+m+1)(k-m-2) *
(n-1)/m(n-1-m) +(n-1)/(k-m-3)( n-k+m+2) *(n-1)/(m+1) (n-m-2) +......
= (n-1)*(n-1) { 1/[(k-m-1)(n-k+m) (m-1)(n-m) ] + 1/[ (n-k+m+1)(k-m-2) m
(n-1-m) ] + 1/[(k-m-3)( n-k+m+2) (m+1) (n-m-2)] +........

taking (m+n-k) and (k-1) common from denominator

finally we will get
1/[(k-1) (m+n-k)]{(n-1) *(n-1+1)(n-1+2)(n-1+3)........(n-1+m)} How does this follow from the previous line?

= 1/[(k-1) (m+n-k)]{(m+n-1))}
=(m+n-1)/[(k-1) (m+n-k)]
= m+n-1Ck-1

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Problem:

I'm trying to prove the following:

using the following identity:

I have tried various methods, but I am baffeled!

Incidentally, I would be very grateful if you could also show me how to prove the second formula.

Solution:

We can solve following identity without using the ...

Solution Summary

Various methods are used to solve trigonometric functions. Left-hand side terms by using the formulas are determined.

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