cos 70°/sin 20° + cos 55° cosec 35°/ tan 5° tan 25° tan 65° tan 85°
Anonymous:
cos madarchod sin madarchod
Answers
Answered by
11
Answer:
Step-by-step explanation:
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 [ans]
Similar questions