trignometry ssc board mathematic part-II standard 10th page 138 question no. V (9) sum
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Using identity a^3-b^3 =(a-b) (a^2+b^2+ab)
So (tan^3@-1)/tan@-1 = (tan@-1) (tan^2@+1^2+tan@*1)/(tan@-1)
By cancelling out tan@-1/tan@-1
We have left only (tan^2@+1^2+tan@*1)
Which is (sec^2@+ tan@).
So (tan^3@-1)/tan@-1 = (tan@-1) (tan^2@+1^2+tan@*1)/(tan@-1)
By cancelling out tan@-1/tan@-1
We have left only (tan^2@+1^2+tan@*1)
Which is (sec^2@+ tan@).
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Answer:
Given, 17+6p=9
Transport 17 to RHS, 6p=9−17
6p=−8
Divide both sides by 6
We will get,
6
6p
=
6
−8
P=
3
−4
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