Question

if tan20 ° = p then prove that ( tan610 ° + tan700 ° ) / ( tan560 ° -tan 470 ° ) = ( 1 - p2 ) / ( p2 + 1 )​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{tan\left(20^{\circ}\right)=p}$

Now,

$\tt{\dfrac{tan\left(560^{\circ}\right)+tan\left(700^{\circ}\right)}{tan\left(610^{\circ}\right)-tan\left(470^{\circ}\right)}}$

$\tt{=\dfrac{tan\left(630^{\circ}-20^{\circ}\right)+tan\left(720^{\circ}-20^{\circ}\right)}{tan\left(540^{\circ}+20^{\circ}\right)-tan\left(450^{\circ}+20^{\circ}\right)}}$

$\tt{=\dfrac{tan\left(7\times90^{\circ}-20^{\circ}\right)+tan\left(8\times90^{\circ}-20^{\circ}\right)}{tan\left(6\times90^{\circ}+20^{\circ}\right)-tan\left(5\times90^{\circ}+20^{\circ}\right)}}$

$\tt{=\dfrac{cot\left(20^{\circ}\right)-tan\left(20^{\circ}\right)}{tan\left(20^{\circ}\right)+cot\left(20^{\circ}\right)}}$

$\tt{=\dfrac{\dfrac{1}{p}-p}{p+\dfrac{1}{p}}}$

$\tt{=\dfrac{1-p^2}{p^2+1}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:tan20\degree = p$

Consider

$\rm :\longmapsto\:\dfrac{tan610\degree + tan700\degree }{tan560\degree - tan470\degree }$

Now,

$\rm :\longmapsto\:tan610\degree$

$\rm \:  =  \: tan(90\degree \times 7 - 20\degree )$

$\rm \:  =  \: cot20\degree$

$\rm \:  =  \: \dfrac{1}{tan20\degree }$

$\rm \:  =  \: \dfrac{1}{p}$

Now,

$\rm :\longmapsto\:tan700\degree$

$\rm \:  =  \: tan(90\degree \times 8 - 20\degree )$

$\rm \:  =  \: - tan20\degree$

$\rm \:  =  \: - p$

Now,

$\rm :\longmapsto\:tan560\degree$

$\rm \:  =  \: tan(90\degree \times 8 + 20\degree )$

$\rm \:  =  \: tan20\degree$

$\rm \:  =  \: p$

Now,

$\rm :\longmapsto\:tan470\degree$

$\rm \:  =  \: tan(90\degree \times 5 + 20)$

$\rm \:  =  \: - cot20\degree$

$\rm \:  =  \: - \dfrac{1}{tan20\degree }$

$\rm \:  =  \: - \dfrac{1}{p}$

So, Substitute all these values, in

$\rm :\longmapsto\:\dfrac{tan610\degree + tan700\degree }{tan560\degree - tan470\degree }$

we get

$\rm \:  =  \: \dfrac{\dfrac{1}{p} - p}{p + \dfrac{1}{p} }$

$\rm \:  =  \: \dfrac{\dfrac{1 - {p}^{2} }{p}}{\dfrac{ {p}^{2} + 1}{p} }$

$\rm \:  =  \: \dfrac{1 - {p}^{2} }{1 + {p}^{2} }$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{tan610\degree + tan700\degree }{tan560\degree - tan470\degree } = \frac{1 - {p}^{2} }{1 + {p}^{2} } }}$

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MORE TO KNOW

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ