Question

secø+tanø=p find cosecø​

Answer 1

$\underline{\underline{\mathfrak{\green{Answer:-}}}}$

cosecA = $\dfrac{{p}^{2}+1}{{p}^{2}-1}$

$\underline{\underline{\mathfrak{\green{Explanation:-}}}}$

Given:

secA + tanA = p -------(1)

To find:

cosecA

Solution:

As we know

${sec}^{2}A - {tan}^{2}A$= 1

(secA+tanA) (secA-tanA) = 1

p(secA-tanA) = 1

secA -tanA =$\dfrac{1}{p}$ -------(2)

Now, eq. (1) +(2)

secA + tanA + secA -tanA = $p+\dfrac{1}{p}$

2secA = $\dfrac{{p}^{2}+1}{p}$

secA = $\dfrac{{p}^{2}+1}{2p}$

$\dfrac{1}{cosA} = \dfrac{{p}^{2}+1}{2p}$

cosA = $\dfrac{2p}{{p}^{2}+1}$

As we know

sinA = $\sqrt{1-{cos}^{2}A}$

sinA = $\sqrt{1-{(\dfrac{2p}{{p}^{2}+1})}^{2}}$

sinA = $\sqrt{1 -\dfrac{4{p}^{2}}{{({p}^{2}+1)}^{2}}}$

sinA = $\sqrt{\dfrac{{({p}^{2}+1)}^{2}-4{p}^{2}}{{({p}^{2}+1)}^{2}}}$

sinA = $\sqrt{\dfrac{{({p}^{2}-1)}^{2}}{{({p}^{2}+1)}^{2}}}$

sinA = $\dfrac{{p}^{2}-1}{{p}^{2}+1}$

$\dfrac{1}{cosecA} = \dfrac{{p}^{2}-1}{{p}^{2}+1}$

cosecA = $\dfrac{{p}^{2}+1}{{p}^{2}-1}$

Answer 2

so the answer is

ANSWER:

$\bf cosecA=\frac{p^2+1}{p^2-1}$

IDENTITIES USED:

→sec²A-tan²A=1

→cosA=1/secA

→sin²A+cos²A=1

→cosecA=1/sinA