Sin 49-cos 41- - cos 17-sin 73- -

The correct option is D 0 Given sin 49^o #160; #160;cos 41^o #160; #160; cos 17^o #160; #160;sin 73 ^o #160; #160; #160; =s ...

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Sum of Trigonometric Ratios in Terms of Their Product

sin 49∘cos 41...

Question

$(\frac{sin 49^{\circ}}{cos 41^{\circ}} - \frac{cos 17^{\circ}}{sin 73^{\circ}}) =$

1

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Solution

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(\frac{\sin 49 °}{\cos 41 °} - \frac{\cos 17 °}{\sin 73 °}) = ?

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

(ii) $c o s 48^{o} - s i n 42^{o}$

(iii) $\frac{c o t 40^{o}}{t a n 50^{o}} - \frac{1}{2} (\frac{c o s 35^{o}}{s i n 55^{o}})$

(iv) ${(\frac{s i n 27^{o}}{c o s 63^{o}})}^{2} + {(\frac{c o s 63^{o}}{s i n 27^{o}})}^{2}$

(v) $\frac{t a n 35^{o}}{t a n 55^{o}} + \frac{c o t 78^{o}}{t a n 12^{o}} - 1$

(vi) $\frac{s e c 70^{o}}{c o s e c 20^{o}} + \frac{s i n 59^{o}}{c o s 31^{o}}$

(vii) $c o s e c 31^{o} - s e c 59^{o}$

(viii) $(s i n 72^{o} + c o s 18^{o}) (s i n 72^{o} - c o s 18^{o})$

(ix) $s i n 35^{o} s i n 55^{o} - c o s 35^{o} c o s 55^{o}$

(x) tan $48^{o} t a n 23^{o} t a n 42^{o} t a n 67^{o}$

(xi) $s e c 50^{o} s i n 40^{o} + c o s 40^{o} c o s e c 50^{o}$

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

$\sum_{r = 1}^{16} \sqrt{1 + c o s 4 r} =$

\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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