Question

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Answer 1

Answer:

tan pie/8 . tan 7pie/8. tan 3pie/8 . tan 5pie/8 [rearranging]

= tan (pie/8) . tan ( 2×pie/2-pie/8) . tan 3pie/2 . tan (2×pie/2-3pie/8) [formula]

tan pie/8 .tan pie/8 . tan 3pie/8 . tan 3pie /8 [minus are cancelled out

=1 (RHS)

Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:tan \bigg(\dfrac{\pi}{8} \bigg)tan \bigg(\dfrac{3\pi}{8} \bigg)tan \bigg(\dfrac{5\pi}{8} \bigg)tan \bigg(\dfrac{7\pi}{8} \bigg) = 1$

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

Consider LHS

$\rm :\longmapsto\:tan \bigg(\dfrac{\pi}{8} \bigg)tan \bigg(\dfrac{3\pi}{8} \bigg)tan \bigg(\dfrac{5\pi}{8} \bigg)tan \bigg(\dfrac{7\pi}{8} \bigg)$

We know,

$\rm :\longmapsto\:\dfrac{\pi}{8} + \dfrac{3\pi}{8} = \dfrac{\pi}{2}$

$\rm :\longmapsto\: \dfrac{3\pi}{8} = \dfrac{\pi}{2} - \dfrac{\pi}{8}$

So,

$\rm :\longmapsto\: tan \bigg(\dfrac{3\pi}{8} \bigg) =tan \bigg( \dfrac{\pi}{2} - \dfrac{\pi}{8} \bigg)$

$\rm :\longmapsto\: tan \bigg(\dfrac{3\pi}{8} \bigg) =cot \bigg(\dfrac{\pi}{8} \bigg)$

$\rm :\longmapsto\: tan \bigg(\dfrac{3\pi}{8} \bigg) =\dfrac{1}{tan \bigg(\dfrac{\pi}{8} \bigg)}$

$\rm :\longmapsto\: tan \bigg(\dfrac{3\pi}{8} \bigg) tan \bigg(\dfrac{\pi}{8} \bigg) = 1 - - - (1)$

Again, we know that,

$\rm :\longmapsto\:\dfrac{5\pi}{8} + \dfrac{7\pi}{8} = \dfrac{3\pi}{2}$

$\rm :\longmapsto\: \dfrac{7\pi}{8} = \dfrac{3\pi}{2} - \dfrac{5\pi}{8}$

So,

$\rm :\longmapsto\: tan \bigg(\dfrac{7\pi}{8} \bigg) =tan \bigg( \dfrac{3\pi}{2} - \dfrac{5\pi}{8} \bigg)$

$\rm :\longmapsto\: tan \bigg(\dfrac{7\pi}{8} \bigg) =cot \bigg(\dfrac{5\pi}{8} \bigg)$

$\rm :\longmapsto\: tan \bigg(\dfrac{7\pi}{8} \bigg) =\dfrac{1}{tan \bigg(\dfrac{5\pi}{8} \bigg)}$

$\rm :\longmapsto\: tan \bigg(\dfrac{5\pi}{8} \bigg) tan \bigg(\dfrac{7\pi}{8} \bigg) = 1 - - - (2)$

Now, Consider

$\rm :\longmapsto\:tan \bigg(\dfrac{\pi}{8} \bigg)tan \bigg(\dfrac{3\pi}{8} \bigg)tan \bigg(\dfrac{5\pi}{8} \bigg)tan \bigg(\dfrac{7\pi}{8} \bigg)$

can be rewritten as using equation (1) and (2) as

$\rm \: = \: \:1 \times 1$

$\rm \: = \: \:1$

Hence,

$\rm :\longmapsto\:tan \bigg(\dfrac{\pi}{8} \bigg)tan \bigg(\dfrac{3\pi}{8} \bigg)tan \bigg(\dfrac{5\pi}{8} \bigg)tan \bigg(\dfrac{7\pi}{8} \bigg) = 1$

Additional Information :-

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