Question

cos10°+sin10°/cos10°-sin10°=cot35°

Answer 1

Hi ,

LHS = ( Cos 10 + Sin 10 ) / ( Cos 10 - Sin 10 )

{ Multiply numerator and and denominator with 1 / Cos 10 }

= [ 1 /Cos 10 ( Cos 10 + Sin 10 ) ] / [ 1/Cos 10 ( Cos 10 - Sin 10 ) ]

= [ Cos10 / Cos10 + Sin10 / Cos10 ] / [ Cos10 / Cos10 - Sin10 / Cos10 ]

= ( 1 + tan10 ) / ( 1 - tan 10 ) [ since sin 10 / cos 10 = tan 10 ]

= ( tan 45 + tan 10 ) / ( 1 - tan 45 tan 10 ) [ since tan 45 = 1 ]

= tan ( 45 + 10 )

[ since tan ( A + B ) = (tanA + tan B ) / (1 - tanAtanB)]

= tan 55

= tan ( 90 - 35 )

= cot 35 [ since tan ( 90 - A ) = cot A ]

= R H S

I hope this helps you.

******

Answer 2

Solution-

LHS-

$\bf \: \frac{cos10\degree + sin10\degree}{cos10\degree - \: sin10\degree}$

$\sf \: \frac{cos10 + sin(90 - 80)}{cos10 - sin(90 - 80)} \\ \\ \sf \: \frac{cos10 + cos80}{cos10 - cos80} \\ \\ \sf \: \frac{cos80 + cos10}{cos10 - cos80} \\ \\ \sf \: \frac{2cos \frac{80 + 10}{2}. \: cos \frac{80 - 10}{2} }{2sin \frac{10 + 80}{2} .sin \frac{80 - 10}{2} } \\ \\ \sf \: cot \frac{80 + 10}{2} \times cot \frac{80 - 10}{2} \\ \\ \sf \: cot \frac{90}{2} \times cot \frac{70}{2} \\ \\ \sf \: cot45 \times cot35 \\ \\ \sf \: 1 \times cot35 \\ \\ \sf \: cot35$