Evaluate -iv tan 35- tan 40- tan 45- tan 50- tan 55-

To evaluate, tan35^∘ , tan40^∘ , tan45^∘ , tan50^∘ , tan55^∘ We know that, tan(90 θ)=θ and θ=1tanθ tan35^∘=tan(90^∘ 55^∘)=55^∘ tan40^∘=t ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

Evaluate :iv ...

Question

Evaluate :
(iv) $tan 35^{\circ} tan 40^{\circ} tan 45^{\circ} tan 50^{\circ} tan 55^{\circ}$

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Solution

To evaluate,

$tan 35^{\circ}, tan 40^{\circ}, tan 45^{\circ}, tan 50^{\circ}, tan 55^{\circ}$

We know that,
$tan (90 - θ) = cot θ$ and $cot θ = \frac{1}{tan θ}$

$tan 35^{\circ} = tan (90^{\circ} - 55^{\circ}) = cot 55^{\circ}$
$tan 40^{\circ} = tan (90^{\circ} - 50^{\circ}) = cot 50^{\circ}$

$∴ tan 35^{\circ} tan 40^{\circ} tan 45^{\circ} tan 50^{\circ} tan 55^{\circ}$

$= cot 55^{\circ} cot 50^{\circ} tan 45^{\circ} tan 50^{\circ} tan 55^{\circ}$

$= \frac{1}{tan 55^{\circ}} \frac{1}{tan 50^{\circ}} tan 45^{\circ} tan 50^{\circ} tan 55^{\circ}$

$= tan 45^{\circ}$

$= 1$

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Similar questions

Find the value of tan35°tan40°tan45°tan50°tan55°

Q. Evaluate :

(i)

\frac{2}{3} (\cos^{4} 30 - \sin^{4} 45 °) - 3 (\sin^{2} 60 ° - \sec^{2} 45 °) + \frac{1}{4} \cot^{2} 30 °

(ii)

4 (\sin^{4} 30 ° + \cos^{4} 60 °) - \frac{2}{3} (\sin^{2} 60 ° - \cos^{2} 45 °) + \frac{1}{2} \tan^{2} 60 °

(iii)

\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 50 ° cosec 40 °

(iv) tan 35° tan 40° tan 45° tan 50° tan 55°
(v) cosec (65° + θ) − sec (25° − θ) − tan (55° − θ) + cot (35° + θ)
(vi) tan 7° tan 23° tan 60° tan 67° tan 83°
(vii)

\frac{2 \sin 68 °}{\cos 22 °} - \frac{2 \cot 15 °}{5 \tan 75 °} - \frac{3 \tan 45 ° \tan 20 ° \tan 40 ° \tan 50 ° \tan 70 °}{5}

(viii)

\frac{3 \cos 55 °}{7 \sin 35 °} - \frac{4 (\cos 70 ° cosec 20 °)}{7 (\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °)}

(ix)

\frac{\sin 18 °}{\cos 72 °} + \sqrt{3} \{\tan 10 ° \tan 30 ° \tan 40 ° \tan 50 ° \tan 80 °\}

(x)

\frac{\cos 58 °}{\sin 32 °} + \frac{\sin 22 °}{\cos 68 °} - \frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °}

(i) $\frac{2}{3} (c o s^{4} 30^{o} - s i n^{4} 45^{o}) - 3 (s i n^{2} 60^{o} - s e c^{2} 45^{o}) + \frac{1}{4} c o t^{2} 30^{o}$

(ii) $4 (s i n^{4} 30^{o} + c o s^{4} 60^{o}) - \frac{2}{3} (s i n^{2} 60^{o} - c o s^{2} 45^{o}) + \frac{1}{2} t a n^{2} 60^{o}$

(iii) $\frac{s i n 50^{o}}{c o s 40^{o}} + \frac{c o s e c 40^{o}}{s e c 50^{o}} - 4 c o s 50^{o} c o s e c 40^{o}$

(iv) $t a n 35^{o} t a n 40^{o} t a n 45^{o} t a n 50^{o} t a n 55^{o}$

(v) $c o s e c (65^{o} + θ) - s e c (25^{o} - θ) - t a n (55^{o} - θ) + c o t (35^{o} + θ)$

(vi) $t a n 7^{o} t a n 23^{o} t a n 60^{o} t a n 67^{o} t a n 83^{o}$

(vii) $\frac{2 s i n 68^{o}}{c o s 22^{o}} - \frac{2 c o t 15^{o}}{5 t a n 75^{o}} - \frac{3 t a n 45^{o} t a n 20^{o} t a n 40^{o} t a n 50^{o} t a n 70^{o}}{5}$
(viii) $\frac{3 c o s 55^{o}}{7 s i n 35^{o}} - \frac{4 (c o s 70^{o} c o s e c 20^{o})}{7 (t a n 5^{o} t a n 25^{o} t a n 45^{o} t a n 65^{o} t a n 85^{o})}$
(ix) $\frac{s i n 18^{o}}{c o s 72^{o}} + \sqrt{3} {t a n 10^{o} t a n 30^{o} t a n 40^{o} t a n 50^{o} t a n 80^{o}}$
(x) $\frac{c o s 58^{o}}{s i n 32^{o}} + \frac{s i n 22^{o}}{c o s 68^{o}} - \frac{c o s 38^{o} c o s e c 52^{o}}{t a n 18^{o} t a n 35^{o} t a n 60^{o} t a n 72^{o} t a n 55^{o}}$

Q. Prove that:

tan 20^{\circ} tan 35^{\circ} tan 45^{\circ} tan 55^{\circ} tan 70^{\circ} = 1

Q. Evaluate :
(vii)

\frac{2 sin 68^{\circ}}{cos 22^{\circ}} - \frac{2 cot 15^{\circ}}{5 tan 75^{\circ}} - \frac{3 tan 45^{\circ} tan 20^{\circ} tan 40^{\circ} tan 50^{\circ} tan 70^{\circ}}{5}

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Evaluate :(iv)tan35∘tan40∘tan45∘tan50∘tan55∘

Evaluate :
(iv) $tan 35^{\circ} tan 40^{\circ} tan 45^{\circ} tan 50^{\circ} tan 55^{\circ}$