Sin 18°/cos72°+3(tan10° tan30° tan40° tan50 °tan80°)
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Heya Mate!!!
☮________$ØLÜTÏØÑ__________☮
= (SIN18°/COS72°) + 3(TAN10° TAN30° TAN40° TAN50° TAN80°)
= ( SIN(90° - 72°) / COS72° ) + 3(TAN( 90° - 80°) (1/√3) TAN( 90° - 50°) × (1/COT50°) × (1/COT80°) )
= (COS72°/COS72°) + 3(COT80° × (1/√3) × COT50° × (1/COT50°) × (1/COT80°)
AFTER SOLVING THIS , WE GET...
=1 + 3 ( 1/√3)
= 1+ √3
Hope it helps uh a lot...
☮________$ØLÜTÏØÑ__________☮
= (SIN18°/COS72°) + 3(TAN10° TAN30° TAN40° TAN50° TAN80°)
= ( SIN(90° - 72°) / COS72° ) + 3(TAN( 90° - 80°) (1/√3) TAN( 90° - 50°) × (1/COT50°) × (1/COT80°) )
= (COS72°/COS72°) + 3(COT80° × (1/√3) × COT50° × (1/COT50°) × (1/COT80°)
AFTER SOLVING THIS , WE GET...
=1 + 3 ( 1/√3)
= 1+ √3
Hope it helps uh a lot...
Sakshibatra51:
Is this write?
Explain how 3(tan(90°-50°)
its right
tan(90°-50°) = Cot50°
tan(90° - ø) = cotø
Thank u I got it
Ur welcome
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