Question

if secA+tanA=p,then find value of cosecA

Answer 1

Answer:

Step-by-step explanation:

secA+tanA=p ........................1

sec²A=1+tan²A (trignometric identity)

∴sec²A-tan²A=1

⇒(secA+tanA)(secA-tanA)=1    [a²-b²=(a+b)(a-b)]

⇒p(secA-tanA)=1   (from .......1)

⇒secA-tanA=1/p......................2

Now we have two equations: ........1 and...........2.

On adding .......1 and ........2 , we see that tanA gets cancelled and we get,

2secA=p+1/p

⇒2secA=(p²+1)/p

⇒secA=(p²+1)/2p

∴cosA=2p/(p²+1)       [∵ cosA= 1/secA]

Now Consider a right angled Δ,right angled at B.

we know that cosA=adjacent side/hypotenuse

∴AB=2p     and AC=p²+1       Now we have to find BC , which we can find on applying pythoguras thearm:

AC²=AB²+BC²     (PYTHOGURAS)

BC²=AC²-AB²

=(p²+1)²-(2p)²=p^4+1+2p²-4p²=p^4+1-2p²=(p²-1)²

∴BC²=(p²-1)²

⇒BC=p²-1

∵cosecA=AC/BC

=(p²+1)/(p²-1)

CosecA=(p²+1)/(p²-1)

Answer 2

$\huge\underline\mathfrak{Answer}$

$\csc( \alpha ) = \frac{ {p}^{2} + 1 }{ {p}^{2} - 1}$

$\huge\underline\mathfrak{Solution}$

$\textbf{Identity,used}$

${ \sec }^{2} \alpha - { \tan }^{2} \alpha = 1$

$\textbf{Step-by-step- Solution}$

${ \sec }^{2} \alpha - { \tan }^{2} \alpha = 1$

$( \sec( \alpha ) - \tan( \alpha ) )( \sec( \alpha ) + \tan( \alpha ) ) = 1$

Given

$\sec( \alpha ) + \tan( \alpha ) = p$

$\textbf{So,}$

$p \times ( \sec( \alpha ) - \tan( \alpha )) = 1$

Eq (1)

$\sec( \alpha ) - \tan( \alpha ) = \frac{1}{p}$

EQ (2)

$\sec( \alpha ) + \tan( \alpha ) = p$

so Substract eq (2) from (1) we , get

$2 \tan( \alpha ) = p - \frac{1}{p}$

$\tan( \alpha ) = \frac{ {p}^{2} - 1 }{2p}$

so ,

$\cot( \alpha ) = \frac{2p}{ {p}^{2} - 1 }$

$\textbf{Using identity}$

${ \csc}^{2} \alpha - { \cot }^{2} \alpha = 1$
so,

${ cosec }^{2} \alpha = 1 + { \cot}^{2} \alpha$
put Value. of Cot (alpha )

${ \csc }^{2} \alpha = 1 + { \: (\frac{2p}{ {p}^{2} - 1} )}^{2}$

${ \csc }^{2} \alpha = \frac{ {( {p}^{2} - 1 )}^{2} + {(2p)}^{2} }{ { ({p}^{2} - 1) }^{2} }$

${ \csc }^{2} \alpha = \frac{ { {p}^{4} - 2 {p}^{2} - 1 + 4 {p}^{2} } }{ { ({p}^{2} - 1) }^{2} }$

${ \csc }^{2} \alpha = \frac{ {( {p}^{2} + 1)}^{2} }{ { ({p}^{2} - 1)}^{2} }$

$\csc( \alpha ) = \frac{ {p}^{2} + 1 }{ {p}^{2} - 1}$

$\huge\underline\mathfrak{Hence}$

$<marquee>$

$\csc( \alpha ) = \frac{ {p}^{2} + 1 }{ {p}^{2} - 1}$