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# 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°).

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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,

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1. Solution:

Given,

(i) (sin 65o/ cos 25o) + (cos 32o/sin 58o) – sin 28o sec 62o + cosec2 30o

= (sin 65o/ cos (90o – 65o)) + (cos 32o/sin (90o – 32o)) – sin 28o sec (90o – 28o) + 22

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

= 1 + 1 – 1 + 4

= 5

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

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

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

= 1 + 2[(cot 8° tan 8o) (cot 17° tan 17o) cot 45°] – 3(1)

= 1 + 2[1 x 1 x 1] – 3

= 1 + 2 – 3

= 0

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