Math, asked by Anonymous, 1 year ago

hey guys please answer my question with proper steps and I'll mark u brainliest

Attachments:

Anonymous: sexxxy
musaib0: hi

Answers

Answered by ranjanalok961
1
hey guys please answer my question with proper steps and I'll mark u brainliest
Attachments:

ranjanalok961: see again hope now you understand clearly .
Anonymous: boobs
ranjanalok961: for find hole value of answer in term of p , so i find value of cisA
ranjanalok961: we find , in another page see i attached there cos = 2p/p²+1
ranjanalok961: its ok
Answered by Anonymous
1
sec + tan = p

As sec^2 - tan^2 = 1

sec+ tan )( sec - tan ) = 1

sec - tan = 1/p

So add both

2 sec = p + 1/p

sec = p + 1/p)/2

1/cos = p + 1/p)/2

1/√1 - sin^2 = p + 1/p)/2

squaring

1/1- Sin^2 = ( p+ 1/p)^2/4

4/ ( p^2 + 1/p^2 + 2) = 1 - sin^2

sin^2 = 1 - 4/ ( p^2 + 1/p^2 +2)

sin^2 = P^2 + 1/p^2 + 2 - 4)/ p^2 + 1/p^2 +2

= p^2 + 1/p^2 -2)/ ( p+ 1/p)^2

= (p -1/p)^2/ ( p+ 1/p)^2

sin A =+- ( p - 1/p)/ p +1/p

= p^2 - 1)/ p^2 +1

cosec A = p^2 +1)/ p^2 -1
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