Simplify cos 80-sin 10- - cos 59- cosec 31-

cos 80 #176; #160; sin 10 #176; #160; +cos 59 #176; cosec31 #176;=cos( 90 #176; 10 #176; ) #160; #160; sin 10 #176; #160; +c ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Simplify co...

Question

Simplify $\frac{cos 80^{\circ}}{sin 10^{\circ}} + cos 59^{\circ} c o s e c 31^{\circ}$

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Solution

$\frac{cos 80 °}{sin 10 °} + cos 59 ° c o s e c 31 ° = \frac{cos (90 ° - 10 °)}{sin 10 °} + cos (90 ° - 31 °) c o s e c 31 °$

$= \frac{sin 10 °}{sin 10 °} + sin 31 ° c o s e c 31 °$

$= 1 + 1$

$= 2.$

Hence, the answer is $2.$

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Similar questions

Q. Find the values of the following.
(i) cos 38

°

cos 52

°

-

sin 38

°

sin 52

°

(ii)

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

(iii)

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

Prove that:

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

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

(iii) $\frac{2 s i n 68^{\circ}}{c o s 22^{\circ}} - \frac{2 c o t 15^{\circ}}{5 t a n 75^{\circ}} - \frac{3 t a n 45^{\circ} t a n 20^{\circ} t a n 40^{\circ} t a n 50^{\circ} t a n 70^{\circ}}{5} = 1$

(iv) $\frac{s i n 18^{\circ}}{c o s 72^{\circ}} + \sqrt{3} (t a n 10^{\circ} t a n 30^{\circ} t a n 40^{\circ} t a n 50^{\circ} t a n 80^{\circ}) = 2$

(v) $\frac{7 c o s 55^{\circ}}{3 s i n 35^{\circ}} - \frac{4 (c o s 70^{\circ} c o s e c 20^{\circ})}{3 (t a n 5^{\circ} t a n 25^{\circ} t a n 45^{\circ} t a n 65^{\circ} t a n 85^{\circ})} = 1$

Prove that $sin 80^{\circ} cos 10^{\circ}$ + $cos 80^{\circ} s i n 10^{\circ}$ = 1

Show that $s i n 10 ° - c o s 10 ° s i n 80 ° c o s 80 ° = 1$

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Simplify cos80∘sin10∘+cos59∘cosec31∘

cos80°sin10°+cos59°cosec31°=cos(90°−10°)sin10°+cos(90°−31°)cosec31° =sin10°sin10°+sin31°cosec31° =1+1 =2.Hence, the answer is 2.

Simplify $\frac{cos 80^{\circ}}{sin 10^{\circ}} + cos 59^{\circ} c o s e c 31^{\circ}$

$\frac{cos 80 °}{sin 10 °} + cos 59 ° c o s e c 31 ° = \frac{cos (90 ° - 10 °)}{sin 10 °} + cos (90 ° - 31 °) c o s e c 31 °$

$= \frac{sin 10 °}{sin 10 °} + sin 31 ° c o s e c 31 °$

$= 1 + 1$

$= 2.$

Hence, the answer is $2.$