Question

find the value of tan 27 degree​

Answer 1

Answer:

Solution:

We have, (sin 27° + cos 27°)2 = sin2 27° + cos2 27° + 2 sin 27° cos 27°

⇒ (sin 27° + cos 27°)2 = 1+ sin 2 ∙ 27°

⇒ (sin 27° + cos 27°)2 = 1 + sin 54°

⇒ (sin 27° + cos 27°)2 = 1 + sin (90° - 36°)

⇒ (sin 27° + cos 27°)2 = 1 + cos 36°

⇒ (sin 27° + cos 27°)2 = 1+ √5+14

⇒ (sin 27° + cos 27°)2 = 14 ( 5 + √ 5)

Therefore, sin 27° + cos 27° = 125+5–√−−−−−−√ …………….….(i)

[Since, sin 27° > 0 and cos 27° > 0 Similarly, we have,

(sin 27° - cos 27°)2 = 1 - cos 36°

⇒ (sin 27° - cos 27°)2 = 1 - √5+14

⇒ (sin 27° - cos 27°)2 = 14 (3 - √5 )

Therefore, sin 27° - cos 27° = ± 123−5–√−−−−−−√ …………….….(ii)

Now, sin 27° - cos 27° = √2 (1√2 sin 27˚ - 1√2 cos 27°)

=√2 (cos 45° sin 27° - sin 45° cos 27°)

= √2 sin (27° - 45°)

= -√2 sin 18° < 0

Therefore, from (ii) we get,

sin 27° - cos 27° = -123−5–√−−−−−−√ …………….….(iii)

Now, adding (i) and (iii) we get,

2 sin 27° = 125+5–√−−−−−−√ - 123−5–√−−−−−−√

⇒ sin 27° = 14(5+5–√−−−−−−√−3−5–√−−−−−−√)

Therefore, sin 27° = 14(5+5–√−−−−−−√−3−5–√−−−−−−√)…………….….(iv)

Again, subtracting (iii) and (i) we get,

2 cos 27° = 125+5–√−−−−−−√ + 123−5–√−−−−−−√

⇒ cos 27° = 14(5+5–√−−−−−−√+3−5–√−−−−−−√)

Therefore, cos 27° = 14(5+5–√−−−−−−√+3−5–√−−−−−−√)…………….….(v)

Now dividing (iv) by (v) we get,

tan 27° = 5+5√√−3−5√√5+5√√+3−5√√

Answer 2

Answer:

tan 27° = 0.50952

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