Question

Without using trigonometric table evaluate:

Sec37/cosec53 + 2cot15 × cot25 × cot45 × cot75 × cot65 - 3 (sin sq.18 + sin sq.72)

Answer 1

Solution :-

Q: Sec37/cosec53 + 2cot15 × cot25 × cot45 × cot75 × cot65 - 3 (sin²18 + sin²72)

= sec(90 - 53)/cosec53 + 2cot(90 - 75) × cot75 × cot(90 - 65) × cot65 × cot45 - 3(sin²(90 - 72) + sin²72)

= cosec53/cosec53 + 2tan75 × cot75 × tan65 × cot65 × 1 - 3(cos²72 + sin²72)

= 1 + 2 - 3

= 0

Hence,

0 is the required answer.

Answer 2

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Answer:

=> 0

Step-by-step explanation:

Refer the attached picture.