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

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Solution

Tan35.tan40.tan45.tan50.tan55
tan(90-55).tan40.1.tan(90-40).tan55
we know that, tan45 =1,and tan(90-X)=cotx
cot55.tan40.1.cot40.tan55
we know that,cotx=1/tanx
1/tan55.tan40.1.1/tan40.tan55
by solving it we get answer as 1

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Q. Evaluate :
(iv)

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

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

Find the value of: tan 20^o. tan 40^o. tan 50^o. tan 60^o. tan 70^o

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

Tan35.tan40.tan45.tan50.tan55 tan(90-55).tan40.1.tan(90-40).tan55 we know that, tan45 =1,and tan(90-X)=cotx cot55.tan40.1.cot40.tan55 we know that,cotx=1/tanx 1/tan55.tan40.1.1/tan40.tan55 by solving it we get answer as 1

Tan35.tan40.tan45.tan50.tan55
tan(90-55).tan40.1.tan(90-40).tan55
we know that, tan45 =1,and tan(90-X)=cotx
cot55.tan40.1.cot40.tan55
we know that,cotx=1/tanx
1/tan55.tan40.1.1/tan40.tan55
by solving it we get answer as 1