tg(π/4+x)+tg(π/4-x)/ctg(π/4+x)+ctg(π/4-x)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should convert the sums to numerator and denominator into products using the following formulas such that:

`tan a + tan b = (sin(a+b))/(cos a*cos b)`

`cot a + cot b = (sin(a+b))/(sin a*sin b)`

Reasoning by analogy yields:

`tg(pi/4+x)+tg(pi/4-x) = (sin(pi/4+x+pi/4-x))/(cos(pi/4+x)*cos(pi/4-x))`

`tg(pi/4+x)+tg(pi/4-x) = sin (2pi/4)/(sqrt2/2(cos x - sin x)*sqrt2/2(cos x + sin x))`

`tg(pi/4+x)+tg(pi/4-x) = sin (pi/2)/(1/2*(cos^2 x - sin^2 x))`

Substituting 1 for sin `pi/2`  and `cos 2x`  for `cos^2 x - sin^2 x`  yields:

`tg(pi/4+x)+tg(pi/4-x) = 1/((1/2)*cos 2x)`

`tg(pi/4+x)+tg(pi/4-x) = 2/(cos 2x)`

`ctg(pi/4+x)+ctg(pi/4-x) = (sin(pi/4+x+pi/4-x))/(sin (pi/4+x)*sin (pi/4-x))`

`ctg(pi/4+x)+ctg(pi/4-x) = (sin (pi/2))/(sqrt2/2(cos x + sin x)*sqrt2/2(cos x - sin x))`

`ctg(pi/4+x)+ctg(pi/4-x) = 1/(1/2*cos 2x)`

`ctg(pi/4+x)+ctg(pi/4-x) = 2/(cos 2x)`

Hence, substituting `2/(cos 2x)`  for`tg(pi/4+x)+tg(pi/4-x)`  and `ctg(pi/4+x)+ctg(pi/4-x)`  yields:

`(tg(pi/4+x)+tg(pi/4-x))/(ctg(pi/4+x)+ctg(pi/4-x)) = (2/(cos 2x))/(2/(cos 2x))`

`(tg(pi/4+x)+tg(pi/4-x))/(ctg(pi/4+x)+ctg(pi/4-x)) = 1`

Hence, evaluating the given expression yields `(tg(pi/4+x)+tg(pi/4-x))/(ctg(pi/4+x)+ctg(pi/4-x)) = 1.`

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