Solve - sin 70 - - cos 40 -

The correct option is B sin 10 ^ ∘ sin70^∘ cos40^∘ =sin70^∘ cos(90^∘ 50^∘) =sin70^∘ sin50^∘ =2>sin(70^∘ 50^∘/2)>cos(70^∘+50^∘/2) ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

Solve : sin...

Question

Solve : $sin 70^{\circ} - cos 40^{\circ}$

sin 10^{\circ}

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cos 80^{\circ}

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tan 10^{\circ}

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cot 10^{\circ}

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\frac{\cos 80 °}{\sin 10 °} + \cos 59 ° \cos e c 31 °

\frac{2 \tan 53 °}{c o t 37 °} - \frac{\cos 80 °}{\tan 10 °}

Prove that:

(i) tan $20^{o} t a n 35^{o} t a n 45^{o} t a n 55^{o} t a n 70^{o} = 1$

(ii) sin $48^{o} s e c 42^{o} + c o s 48^{o} c o s e c 42^{o} = 2$

(iii) $\frac{s i n 70^{o}}{c o s 20^{o}} + \frac{c o s e c 20^{o}}{s e c 70^{o}} - 2 c o s 70^{o} c o s e c 20^{o} = 0$

(iv) $\frac{c o s 80^{o}}{s i n 10^{o}} + c o s 59^{o} c o s e c 31^{o} = 2$

$c o s 40^{\circ} + c o s 80^{\circ} + c o s 160^{\circ} + c o s 240^{\circ} =$

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