Prove that
Sec41°sin49° + cos29°cosec61° - 2/√3(tan20°tan60°tan30°)/3(sin^31° + sin^59°)=0

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Solution

[sec41°sin49°+cos29°cosec61°-2/√3(tan20°tan60°tan70°)]/[3(sin²31°+sin²59°)]
=[sec41°sin(90°-41°)+cos29°cosec(90°-29°)-2/√3{tan20°×√3×tan(90°-20°)}]/
[3{sin²31°+sin²(90°-31°)]
=[sec41°cos41°+cos29°sec29°-2(tan20°cot20°)]/[3(sin²31°+cos²31°]
=(1+1-2)/(3×1)
=0

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\frac{\tan 60 ° - \tan 30 °}{1 + \tan 60 ° \tan 30 °}

\frac{cos 70^{o}}{sin 20^{o}} + \frac{cos 59^{o}}{sin 31^{o}} - 8 {sin}^{2} 30^{o} = 0

\frac{tan 60^{\circ} - tan 30^{\circ}}{1 + tan 60^{\circ} tan 30^{\circ}}

Prove that:

(i) $t a n 5^{\circ} t a n 25^{\circ} t a n 30^{\circ} t a n 65^{\circ} t a n 85^{\circ} = \frac{1}{\sqrt{3}}$

(ii) $c o t 12^{\circ} c o t 38^{\circ} c o t 52^{\circ} c o t 60^{\circ} c o t 78^{\circ} = \frac{1}{\sqrt{3}}$

(iii) $c o s 15^{\circ} c o s 35^{\circ} c o s e c 55^{\circ} c o s 60^{\circ} c o s e c 75^{\circ} = \frac{1}{2}$

(iv) $c o s 1^{\circ} c o s 2^{\circ} c o s 3^{\circ} . . . c o s 180^{\circ} = 0$

(v) ${(\frac{s i n 49^{\circ}}{c o s 41^{\circ}})}^{2} + {(\frac{c o s 41^{\circ}}{s i n 49^{\circ}})}^{2} = 2$

\frac{1}{\sqrt{3}}

\frac{1}{\sqrt{3}}

\frac{1}{2}

{(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2} = 2

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