Question

if sec0 + tan0= p the find the value of cosec0

Answer 1

(sinA+cosA)X secA= (sinA+cosA)X1/cosA

=sinA/cosA+cosA/cosA

=tanA+1

=5/12+1

=17/12

Answer 2

Answer:

Value of cosec Θ = $\frac{p^{2}+1}{p^{2}-1}$

Step-by-step explanation:

Given,

sec Θ + tan Θ = p   ( Let this be Eqn 1 )

Since we know,

sec² Θ - tan² Θ = 1

We can write it as,

(sec Θ + tan Θ ) ( sec Θ - tan Θ ) = 1

Substituting the value of sec Θ + tan Θ

p ( sec Θ - tan Θ ) = 1

sec Θ - tan Θ = 1 / p   (Let this be Eqn 2 )

Adding Eqn 1 and Eqn 2,

sec Θ + tan Θ + sec Θ - tan Θ  = p + 1 / p

2 sec Θ = p + $\frac{1}{p}$

2 sec Θ = $\frac{p^{2}+1}{p}$

sec Θ = $\frac{p^{2}+1}{2p}$

We know,

cos Θ =  1 / sec Θ , that is,

1 / $\frac{p^{2}+1}{2p}$

cos Θ = $\frac{2p}{p^{2}+1 }$

Since,

sin Θ = √ 1 - cos² Θ

Substituting the value of cos Θ

sin Θ = $\sqrt{1-(\frac{2p}{p^{2}+1 })^{2} }$

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

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

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

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

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

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

So,

sin Θ =  $\frac{p^{2}-1}{p^{2}+1}$

Since,

cosec Θ = 1 /  sin Θ

cosec Θ = 1 /  $\frac{p^{2}-1}{p^{2}+1}$

cosec Θ = $\frac{p^{2}+1}{p^{2}-1}$

Hence,

Value of cosec Θ = $\frac{p^{2}+1}{p^{2}-1}$