Question

Tan10.tan50+tan50.tan70+tan70.tan170=3

Answer 1

Answer:

Tan10.tan50+tan50.tan70+tan70.tan170=3

Step-by-step explanation:

Tan10.tan50+tan50.tan70+tan70.tan170=3

tan170 = -Tan (180 - 170) = - Tan 10

=> LHS

= Tan10.tan50+tan50.tan70+tan70.(-Tan10)

= Tan10 . Tan (60-10) + Tan(60-10).Tan(60+10) - Tan(60+10)Tan10

= Tan10 ( Tan60 - Tan10)/(1 + Tan60Tan10) - Tan10(( Tan60 + Tan10)/(1 - Tan60Tan10) + ( Tan60 - Tan10)/(1 + Tan60Tan10)*( Tan60 + Tan10)/(1 - Tan60Tan10)

= Tan10 (Tan60 - Tan10 - Tan60Tan10(Tan60 - Tan10) - Tan60 - Tan10 -Tan60Tan10(Tan60 + Tan10))/(1-Tan²60Tan10)  + (Tan²60 - Tan²10)/(1  - Tan²60Tan²10)

= Tan10(-2Tan10 -2Tan²60Tan10)  + Tan²60 - Tan²10) /(1  - Tan²60Tan²10)

=  (-2Tan²10 - 2Tan²60Tan²10 + Tan²60 - Tan²10)/(1  - Tan²60Tan²10)

Using Tan²60 = 3

= (-2Tan²10 - 2*3Tan²10 + 3 - Tan²10)/(1  - 3Tan²10)

= (3 - 9Tan²10)/(1  - 3Tan²10)

= 3(1 - 3Tan²10)/(1  - 3Tan²10)

Canceling (1  - 3Tan²10) from numerator & Denominator

= 3

= RHS

QED

Proved

Tan10.tan50+tan50.tan70+tan70.tan170=3