you can find trig laws on google.
tan+sin
(sin/cos+sin)(top first)
(sin/cos+sin/1)
(2sin/(cos+1))
tan-sin
(sin/cos)-sin/1(bottom)
(2sin/(cos-1))
(2sin/(cos+1))/(2sin/(cos-1))(together)
2sin cancels
(cos-1)/(cos+1)(what's left)
right side of the equals sign
sec+1
(1/cos)+1/1 (top)
(2/cos+1)
(1/cos)-1/1(bottom)
(2/cos-1)
2 cancels
(cos-1)/(cos+1)(whats left)
therefore Tan+sin/tan-sin=sec+1/sec-1
Trig is a huge puzzle, so if you get stuck on one thing try a different. Changing to sin and cos could be your friend. All the laws I used were on that page.
Prove that:Tanø+sinø/tanø-sinø=secø+1/secø-1
2 answers
surely you are joking!
sin/cos + sin/1 is certainly NOT 2sin/(cos+1)
any more than 4/3 + 4/1 = 2*4/(3+1) = 2
It just happens that your mistakes eliminated each other.
sin = tan*cos, so
tan+sin = tan(1+cos)
tan-sin = tan(1-cos)
divide and you get (1+cos)/(1-cos)
divide top and bottom by cos and you have
(sec+1)/(sec-1)
sin/cos + sin/1 is certainly NOT 2sin/(cos+1)
any more than 4/3 + 4/1 = 2*4/(3+1) = 2
It just happens that your mistakes eliminated each other.
sin = tan*cos, so
tan+sin = tan(1+cos)
tan-sin = tan(1-cos)
divide and you get (1+cos)/(1-cos)
divide top and bottom by cos and you have
(sec+1)/(sec-1)