Question

Page No
7. Cosec a + cot A =P then
prove that
sec A = P²+1 /p 2 -1

Answer 1

Assuming that all the letters are written with caps lock off ^_^"

We have,

cosec a + cot a = p. ---(i)


Also we have an identity as,

1+ cot² a = cosec² a
Which in furthur shifting of terms give,

cosec² a - cot² a= 1.

Which can also be written as,
(cosec a + cot a ) (cosec a -cot a ) = 1

substituting the value from (i)


p( cosec a - cot a ) = 1

cosec a -cot a = 1/p. -----(ii)


Adding (i) and (ii) gives,


$2cosec \: \: a = p + \frac{1}{p} \\ \\ cosec \: a = \frac{ {p}^{2} + 1 }{2p} \\ \\ \ \sin \: a = \frac{2p}{ {p}^{2} + 1 }$


Now we have nother identity as,

${ \sin }^{2} a + { \cos }^{2} a = 1$

Substituing the value from sin we have,

$\frac{ 4{p}^{2} }{ {( {p}^{2} + 1) }^{2} } + { \cos }^{2}a = 1 \\ \\ { \cos }^{2} a = 1 - \frac{4 {p}^{2} }{ {( {p}^{2} + 1) }^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {( {p}^{2} + 1)}^{2} - 4 {p}^{2} }{ {( {p}^{2} + 1)}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {( {p}^{2} - 1) }^{2} }{ {( {p}^{2} + 1)}^{2} } \\ \\ \cos \: a = \sqrt{ \frac{ {( {p}^{2} - 1)}^{2} }{ {( {p}^{2} + 1)}^{2} } } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ {p}^{2} - 1}{ {p}^{2} + 1 }$


Which implies that,

$\sec \: a = \frac{ {p}^{2} + 1}{ {p}^{2} - 1 }$


Hence your answer.

Hope my efforts were helpful to you.
^_^

Answer 2

Step-by-step explanation:

(cosec a+cota) square +1

-----------------------------------

(cosec a+cota) square-1

cosec square a+ cot square a+2cosec a

cot a +1

--------------------------------------------------------

cosec square a +cot square a +2cosec a

cot a -1

cosec square a+ cosec square a+2cosec a

-------------------------------------------------------------

cot square a + cot squarea +2 cosec a

since 1+ cot squarea= cosec square a

2cosec a(cosec a+1)

_______________

2(cot square a+ cosec a)

1/sina(1/sina+1)

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cos squarea/sin square a+1/sina

1/sina(1+sina)/sina

-------------------------

(cos square a+sina)/ sin square a