If tan theta + sec theta=x, show that 2tan theta =x-1÷x, 2 sec theta=x+1÷x.Hence show that sin theta =x^2-1÷x^2+1
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sec@ + tan@ = x ———(1)
sec@ – tan@ = 1/x ——(2)
add 1&2
2sec@ = x+1/x = (x²+1)/x ——(3)
sub 2 from 1
2tan@ = x–1/x = (x²–1)/x ——(4)
(4)÷(3)
2tan@÷2sec@ = [(x²–1)/x] ÷[(x²+1)/x]
sin@ = (x²–1)/(x²+1)
sec@ – tan@ = 1/x ——(2)
add 1&2
2sec@ = x+1/x = (x²+1)/x ——(3)
sub 2 from 1
2tan@ = x–1/x = (x²–1)/x ——(4)
(4)÷(3)
2tan@÷2sec@ = [(x²–1)/x] ÷[(x²+1)/x]
sin@ = (x²–1)/(x²+1)
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