Prove the trig identities

Question

3, 9 and 7. Use the above numbers to prove only three of these trig identities.

sin x/ 1 + cos x = 1 - cos x/sin x

1/ 1 + Cos x = csc^2 x - cot x/sinx x

-sin x - cos x cot x= - csc x

1 + tan x/ 1 + cot x = tan x

Answer 1

#1)

sin^2 + cos^2 = 1 <-- the MAIN pythagorean identity

sin^2 = 1 - cos^2 <--- subtracts cos^2 from both sides

sin^2 = (1+cos)(1-cos) <--- factors right side

sin/(1+cos) = (1-cos)/sin <--- divides both sides by sin (1+cos)

#2) csc^2 - cot/sin =

csc^2 - (cos/sin)(1/sin) <--- cot = cos/sin; other sine is in the denominator

csc^2 - cos/sin^2 <--- multiplies the fractions: cos *1 = cos, sin*sin=sin^2

=1/sin^2 - cos/sin^2 <-- csc = 1/sin, so csc^2 = 1/sin^2

=(1 - cos)/sin^2 <--- combines into a single fraction since the denominators are both sin^2

=(1-cos)/(1-cos^2) <--- sin^2 = 1 - cos^2 per main pythagorean identity

= (1-cos)/ [(1+cos)(1-cos)] <--- factors denominator (1-cos^2) = (1+cos)(1-cos)

1/(1+cos) <--- 1 - cos cancels

#3)

-sin - cos*cos/sin =

-sin^2/sin - cos*cos/sin <--- common denominator is sin; so 1st fraction becomes -sin^2

(-sin^2-cos^2)/sin = <--- cos*cos=cos^2; combines into a single fraction per LCD

-(sin^2+cos^2)/sin = <---factors out -1 in numerator

-1/sin = <--- sin^2+cos^2=1 per main pythagorean identity

-csc <--- csc = 1/sin, so -1/sin = -csc

#4) multiplies everything by tangent...

(tan + tan^2)/ (tan + 1) =

tan( 1+ tan)/(tan+1) = <--- factors out tangent in numerator

tan <--- 1+tan cancels