How to solve the 10th question?
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Put x=a × tanФ
Then our question becomes:
Tan-1[(3a³tanФ-a³tan³Ф)/a³-3a³tan²Ф)]
By taking a³ common from numerator and denominator we get
Tan-1[a³{3tanФ-tan³Ф}/a³{1-3tan²Ф}]
a³ gets canceled .
Tan-1[{3tanФ-tan³Ф}/{1-3tan²Ф}]............................(1)
This is in the form of :
Tan 3Ф=(3tanФ-tan³Ф)/1-3tan²Ф [trigonometric formula]
∴in equation (1): instead of [ (3tanФ-tan³Ф)/1-3tan²Ф] put (Tan 3Ф)
∴eqation (1) becomes
Tan-1(tan 3Ф) [tan-1 and tan gets canceled]
=3Ф....................... (2)
Now we know x =a tan Ф
x/a=tanФ
tan-1(x/a) = Ф
Putting the value of Фin eqn (2)
We get
3× tan-1(x/a),which is our answer.... :)
Tan-1[(3a²x-x³)/a³-3ax² = 3× tan-1(x/a)
Then our question becomes:
Tan-1[(3a³tanФ-a³tan³Ф)/a³-3a³tan²Ф)]
By taking a³ common from numerator and denominator we get
Tan-1[a³{3tanФ-tan³Ф}/a³{1-3tan²Ф}]
a³ gets canceled .
Tan-1[{3tanФ-tan³Ф}/{1-3tan²Ф}]............................(1)
This is in the form of :
Tan 3Ф=(3tanФ-tan³Ф)/1-3tan²Ф [trigonometric formula]
∴in equation (1): instead of [ (3tanФ-tan³Ф)/1-3tan²Ф] put (Tan 3Ф)
∴eqation (1) becomes
Tan-1(tan 3Ф) [tan-1 and tan gets canceled]
=3Ф....................... (2)
Now we know x =a tan Ф
x/a=tanФ
tan-1(x/a) = Ф
Putting the value of Фin eqn (2)
We get
3× tan-1(x/a),which is our answer.... :)
Tan-1[(3a²x-x³)/a³-3ax² = 3× tan-1(x/a)
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