Math, asked by DilpreetSingh0, 1 year ago

tana+tan(60+a)-tan(60-a)=3tan3a

Answers

Answered by anustarnoor

70

LHS= tan(A) + tan(A+60) - tan(60-A)= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) - (tan 60 - tan(A))/(1 + tan60tan(A))= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) + (tan A - tan 60)/(1 + tan60tan(A))= tan(x) + (tan(x) + √3)/(1 - √3tan(x)) + (tan(x) - √3)/(1 + √3tan(x))= tan(x) + (tan(x) - √3)/(1 + √3tan(x)) + (tan(x) + √3)/(1 - √3tan(x))= tan(x) + [(tan(x) - √3)(1 - √3tan(x)) + (tan(x) + √3)(1 + √3tan(x))]/(1-3tan²(x))= tan(x) + [tan(x) - √3tan²(x) - √3 + 3tan(x) + tan(x) + √3tan²(x) + √3 + 3tan(x)]/(1-3tan²(x))= tan(x) + 8tan(x)/(1-3tan²(x))= (tan(x) - 3tan³(x) + 8tan(x))/(1-3tan²(x))= (9tan(x) - 3tan³(x))/(1-3tan²(x))= 3(3tan(x) - tan³(x))/(1-3tan²(x))= 3tan(3x)= RHS

DilpreetSingh0: thanks very much

DilpreetSingh0: bro typing is difficult to do send images to people.that help much more

Answered by veenitkumar4

14

LHS= tan(A) + tan(A+60) - tan(60-A)= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) - (tan 60 - tan(A))/(1 + tan60tan(A))= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) + (tan A - tan 60)/(1 + tan60tan(A))= tan(x) + (tan(x) + √3)/(1 - √3tan(x)) + (tan(x) - √3)/(1 + √3tan(x))= tan(x) + (tan(x) - √3)/(1 + √3tan(x)) + (tan(x) + √3)/(1 - √3tan(x))= tan(x) + [(tan(x) - √3)(1 - √3tan(x)) + (tan(x) + √3)(1 + √3tan(x))]/(1-3tan²(x))= tan(x) + [tan(x) - √3tan²(x) - √3 + 3tan(x) + tan(x) + √3tan²(x) + √3 + 3tan(x)]/(1-3tan²(x))= tan(x) + 8tan(x)/(1-3tan²(x))= (tan(x) - 3tan³(x) + 8tan(x))/(1-3tan²(x))= (9tan(x) - 3tan³(x))/(1-3tan²(x))= 3(3tan(x) - tan³(x))/(1-3tan²(x))= 3tan(3x)= RHS

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