Math, asked by tanishka200629, 1 day ago

trignometry ssc board mathematic part-II standard 10th page 138 question no. V (9) sum

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Answered by rawat999
0
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@).
Answered by Prince063867
0

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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