Math, asked by SHIKHZCHIKZ4082, 1 year ago

Cos 70÷sin20 + cos 55÷ tan 5tan25 × cosec 35÷ tan65 tan85

Answers

Answered by Anonymous
1

ANSWER:-------

{Cos70°/sin20° + (cos55°.cosec35°)/(tan5°.tan25°.tan65°.tan85°) 

we know, sin(90 - θ) = cosθ 

so, sin20° = sin(90° -70°) = cos70° ----(1)

cos55° = cos(90° - 35°) = sin35° ------(2)

also tan(90° - θ) = cotθ 

so, tan25° = tan(90° - 65°) = cot65° -----(3)

tan5° = tan(90° - 85°) = cot85° -----(4)

from equations (1), (2), (3) and (4), 

= cos70°/cos70° + {sin35°. cosec35°}/{cot85°.cot65°.tan45°.tan65°.tan85°}

= 1 + {sin35° × 1/sin35°}/{(tan85°.cot85°) × tan45° × (tan65°.cot65°)}

= 1 + 1/tan45°

= 1 + 1/1 = 1 + 1 = 2

hope it helps:)

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