Question

Evaluate the following :
(i)$\frac{sin20^{0} }{cos70^{0} }$
(ii)$\frac{cos19^{0} }{sin71^{0} }$
(iii)$\frac{sin21^{0} }{cos69^{0} }$
(iv)$\frac{tan10^{0} }{cot80^{0} }$
(v)$\frac{sec11^{0} }{cosec79^{0} }$

Answer 1

(i) SOLUTION :  

Given : sin 20° /cos 70°

sin 20° /cos 70° = sin(90°−70°)/ cos 70°

sin 20° /cos 70° = cos 70° /  cos 70°  

[ sin (90 – θ) = cos θ]

sin 20° /cos 70° =1

Hence,  sin 20° /cos 70° = 1

(ii) SOLUTION :  

Given : cos 19°/ sin 71°

cos 19°/ sin 71°  = cos(90° −71°)/sin 71°

cos 19°/ sin 71° = sin 71°/sin 71°

[ cos(90 - θ)= sin θ]

cos 19° / sin 71° =1

Hence, cos 19° / sin 71° = 1  

(iii) SOLUTION :  

Given :  sin 21° / cos 69°

sin 21° / cos 69° = sin(90° − 69°) /cos 69°

sin 21° / cos 69° = cos 69°/cos 69°

[sin (90 – θ) = cos θ]

sin 21° / cos 69° = 1

Hence,   sin 21° / cos 69° = 1

(iv) SOLUTION :  

Given : tan 10°/cot 80°

tan 10°/cot 80°  = tan(90°− 80°)/cot 80°

tan 10°/cot 80° = cot 80° /cot 80°

[tan (90 – θ) = cot θ]

tan 10°/cot 80°  = 1

Hence,  tan 10°/cot 80° = 1

(v) SOLUTION :  

Given : sec 11° / sec 79°

sec 11° / sec 79° = sec(90°−79°)/ cosec 79°

sec 11° / sec 79° = cosec 79°/ sec 79°

[ sec(90 - θ) = cosec θ]

sec 11° / sec 79° = 1

Hence,  sec 11° / sec 79° = 1

HOPE THIS ANSWER WILL HELP YOU….  

Answer 2

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