This ques has been asked in 2015 so it is very important question

This question is from the Book- ML Aggarwal

Board- ICSE

Publication- Avichal

Chapter- Trigonometric Identities

Chapter number-18

Without using trigonometric tables, evaluate the following (i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o (ii) (sin 29o/ cosec 61o) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).

Question no. 7th, ML Aggarwal, Class10, chapter 18,

Solution:Given,

(i) (sin 65

^{o}/ cos 25^{o}) + (cos 32^{o}/sin 58^{o}) – sin 28^{o}sec 62^{o}+ cosec^{2}30^{o}= (sin 65

^{o}/ cos (90^{o}– 65^{o})) + (cos 32^{o}/sin (90^{o}– 32^{o})) – sin 28^{o}sec (90^{o}– 28^{o}) + 2^{2}= (sin 65

^{o}/sin 65^{o})^{ }+ (cos 32^{o}/cos 32^{o}) – [sin 28^{o}x cosec 28^{o}] + 4= 1 + 1 – 1 + 4

= 5

(ii) (sin 29

^{o}/ cosec 61^{o}) + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin² 38° + sin² 52°).= (sin 29

^{o}/ cosec (90^{o}– 29^{o})) + [2 cot 8° cot 17° cot 45° cot (90° – 17^{o}) cot (90^{o}– 8^{o})] – 3(sin² 38° + sin² (90° – 38^{o}))= (sin 29

^{o}/ sin 29^{o}) + [2 cot 8° cot 17° cot 45° tan 17^{o}tan 8^{o}] – 3(sin² 38° + cos² 38°)= 1 + 2[(cot 8° tan 8

^{o}) (cot 17° tan 17^{o}) cot 45°] – 3(1)= 1 + 2[1 x 1 x 1] – 3

= 1 + 2 – 3