Question

what is the value of tan 72°​

Answer 1

Answer:

Let, A = 18°

Therefore, 5A = 90°

⇒ 2A + 3A = 90˚

⇒ 2A = 90˚ - 3A

Taking sine on both sides, we get

sin 2A = sin (90˚ - 3A) = cos 3A

⇒ 2 sin A cos A = 4 cos3 A - 3 cos A

⇒ 2 sin A cos A - 4 cos3 A + 3 cos A = 0

⇒ cos A (2 sin A - 4 cos2 A + 3) = 0 Dividing both sides by cos A = cos 18˚ ≠ 0, we get

⇒ 2 sin A - 4 (1 - sin2 A) + 3 = 0

⇒ 4 sin2 A + 2 sin A - 1 = 0, which is a quadratic in sin A

Therefore, sin A = −2±−4(4)(−1)√2(4)

⇒ sin A = −2±4+16√8

⇒ sin A = −2±25√8

⇒ sin A = −1±5√4

Now sin 18° is positive, as 18° lies in first quadrant.

Therefore, sin 18° = sin A = √5−14

Now, cos 72° = cos (90° - 18°) = sin 18° = √5−14

And cos 18° = √(1 - sin2 18°), [Taking positive value, cos 18° > 0]

⇒ cos 18° = 1−(5√−14)2−−−−−−−−−−√

⇒ cos 18° = 16−(5+1−25√)16−−−−−−−−−−√

⇒ cos 18° = 10+25√16−−−−−−√

Thus, sin 72° = sin (90° - 18°) = cos 18° = 10+25√√4

Now, tan 72° = sin72°cos72° = 10+25√√4√5−14 = 10+2√5√√5−1

Therefore, tan 72° =10+2√5√√5−1

Answer 2

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Let, A = 18°

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Therefore, 5A = 90°

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⇒ 2A + 3A = 90˚

⇒ 2A = 90˚ - 3A

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Taking sine on both sides, we get

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sin 2A = sin (90˚ - 3A) = cos 3A

⇒ 2 sin A cos A = 4 cos³ A - 3 cos A

⇒ 2 sin A cos A - 4 cos³ A + 3 cos A = 0

⇒ cos A (2 sin A - 4 cos² A + 3) = 0

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Dividing both sides by cos A = cos 18˚ ≠ 0, we get

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⇒ 2 sin A - 4 (1 - sin² A) + 3 = 0

⇒ 4 sin² A + 2 sin A - 1 = 0, which is a quadratic in sin A

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Therefore, sin A = $\frac{2 +\sqrt{- 4(4)( - 1)} }{2(4)}$

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⇒ sin A = $\frac{ - 2 + \sqrt{4 + 16}}{8}$

⇒ sin A = $\frac{ - 2 + 2 \sqrt{5} }{8}$

⇒ sin A = $\frac{ - 1 + \sqrt{5}}{4}$

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Now sin 18° is positive, as 18° lies in first quadrant.

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Therefore, sin 18° = sin A = $\frac{√5 - 1}{4}$

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Now, cos 72° = cos (90° - 18°) = sin 18° = $\frac{√5 - 1}{4}$

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And cos 18° = √(1 - sin² 18°), [Taking positive value, cos 18° > 0]

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⇒ cos 18° = $\sqrt{1 - \frac{(√5 - 1)}{(4)} 2}$

⇒ cos 18° = $\sqrt\frac{16 - (5 + 1 - 2 \sqrt{5)} }{16}$

⇒ cos 18° = $\sqrt{10 + 2 \sqrt{5} } \\ \frac{}{16}$

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Thus, sin 72° = sin (90° - 18°) = cos 18° = $\sqrt{10 + 2 \sqrt{5}}\\\frac{}{4}$

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Now, tan 72° = $\frac{ \sin72°}{ \cos72°}$ = $\sqrt\frac{10 + 2 \sqrt{5} }{4} \\ \frac{√5 - 1}{4}$ = $\sqrt \frac{10 + 2 \sqrt{5} }{√5 - 1}$

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Therefore, the exact value tan 72° = $\sqrt\frac{10 + 2 \sqrt{5} }{4} \\\frac{}{√5 - 1}$