Question

please help with the answer.​

Answer 1

Answer:

tan0+ sec0=l

tan0=l-sec0......a

now

sec0=l-tan0

sec0= l-(l-sec0)

sec0=l-l+sec0

sec0=sec0

Answer 2

Answer:

$sec\theta = \frac{l^{2} + 1}{2l}$

Step-by-step explanation:

given:

tanθ + secθ = $l$

now take square on both sides,

( tanθ + secθ )^2 = $l^{2}$

$tan^{2} \theta + sec^{2}\theta+2tan\theta sec\theta = l^{2}$

we know that,

$sec^{2}\theta - tan^{2}\theta = 1$    ⇒ $tan^{2}\theta = sec^{2}\theta -1$

so,

$tan^{2} \theta + sec^{2}\theta+2tan\theta sec\theta = l^{2}$     ⇒    $sec^{2}\theta - 1 + sec^{2}\theta + 2 tan\theta sec\theta = l^{2}$

 ⇒   $2 sec^{2}\theta+2 tan\theta sec\theta - 1 = l^{2}$

⇒  $2sec\theta( sec\theta + tan\theta) - 1 = l^{2}$

⇒ $2sec\theta( l ) - 1 =l^{2}$              since   tanθ + secθ = $l$

⇒ $l^{2} - 2sec\theta l + 1 = 0$

 ⇒ $2sec\theta l =l^{2} + 1$

⇒ $sec\theta = \frac{l^{2} + 1}{2l}$

thus you got the answer !