Question

tan9°- tan27°- tan63°+tan81°=?​

Answer 1

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☞ Your Answer is 4

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$\huge\sf\blue{Given}$

✭ $\sf tan 9^{\circ} - tan 27^{\circ} - tan 63^{\circ} + tan 81^{\circ}$

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$\huge\sf\gray{To \:Find}$

◈ The final value?

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➳ $\sf tan 9^{\circ}+tan 81^{\circ} -tan 27^{\circ}-tan63^{\circ}$

➳ $\sf tan ^{\circ}+cot 9^{\circ}-( tan 27^{\circ} + cot 27^{\circ}$

$\bigg\lgroup \sf tan A + cot A = 2 cosec 2A\bigg\rgroup$

➳ $\sf 2 cosec 2(9^{\circ})-(2 cosec 2 (27^{\circ}))$

➳ $\sf 2 cosec 18^{\circ} - 2 cosec 54{\circ}$

➳ $\sf 2 \dfrac{1}{sin 18^{\circ}} - 2 \dfrac{1}{sin 54^{\circ}}$

➳ $\sf 2 \times \dfrac{4}{\sqrt{5}-1} - 2\times \dfrac{4}{\sqrt{5}+1}$

➳ $\sf 8\bigg\lgroup \dfrac{\sqrt{5}+1-(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\bigg\rgroup$

➳ $\sf 8\bigg\lgroup \dfrac{\sqrt{5}+1-\sqrt{5}+1}{5-1}\bigg\rgroup$

➳ $\sf 8\bigg\lgroup \dfrac{2}{4}\bigg\rgroup$

➳ $\sf \orange{tan 9^{\circ}+tan 81^{\circ} -tan 27^{\circ}-tan63^{\circ = 4}}$

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Answer 2

Answer:

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$<font color="blue">$ your answer is 4

Step-by-step explanation:

subsititute tan81 = cot9 and tan63=cot27

By simplifying this you’ll get

tan^2(9) + 1/ tan9 - tan^2(27) +1/ tan27

tan^3(x)+1 = sec^2(x)

siimplify using this and you get

2/sin18 - 2/sin54

this gives you 2*( sin54 - sin18 / sin54sin18 )

= 2 * (2cos36sin18/sin54sin18) { using sinA - sinB = 2 cos(A+B/2)sin(A-B/2) }

= 4 ( cos36sin18/sin54sin18)

= 4

because cos36 = sin54 as 54 and 36 are complementary angles.