Question

If sec A + tan A = p, then find the value of cos A and sin A in terms of p.​

Answer 1

Answer:

1+sin a =p cos a

Step-by-step explanation:

sec a =1/cos a and tan a =sin a /cos a..

lcm is cos a then 1 +sin a / cos a = p taking cos a to the rhs we get 1+ sin a = p cos a

Answer 2

Answer:-

Given:

sec A + tan A = p -- equation (1).

We know that,

sec² A - tan² A = 1

using a² - b² = (a + b)(a - b) we get,

⟹ (sec A + tan A) (sec A - tan A) = 1

Putting the value of sec A + tan A from equation (1) we get,

⟹ p * (sec A - tan A) = 1

⟹ sec A - tan A = 1/p -- equation (2)

Add equations (1) & (2).

$\implies \sf \: \sec A + \tan \: A + \sec \: A \: - \tan \: A = p + \frac{1}{p} \\ \\ \implies \sf \: 2 \: \sec A = \frac{ {p}^{2} + 1 }{p} \\ \\\implies \sf \sec A = \frac{ {p}^{2} + 1 }{p} \times \frac{1}{2} \\ \\ \implies \sf \red{ \sec \: A = \frac{ {p}^{2} + 1}{2p} }$

Substitute the value of sec A in equation (1).

$\implies \sf \frac{ {p}^{2} + 1 }{2p} + \tan \: A = p \\ \\ \implies \sf \tan \: A = p - \frac{p ^{2} + 1 }{2p} \\ \\ \implies \sf \tan \: A = \frac{2 {p}^{2} - ( {p}^{2} + 1) }{2p} \\ \\ \implies \sf \tan \: A = \frac{2 {p}^{2} - {p}^{2} - 1 }{2p} \\ \\ \implies \sf \red{ \tan \: A = \frac{ {p}^{2} - 1}{2p} }$

Divide sec A by tan A.

$\implies \sf \frac{ \sec \: A}{ \tan \: A} = \dfrac{ \frac{ {p}^{2} + 1 }{2p} }{ \frac{ {p}^{2} - 1}{2p} }$

using sec A = 1/cos A & tan A = sin A/cos A we get,

$\: \implies \sf \: \frac{ \frac{1}{ \cos \: A } }{ \frac{ \sin \: A}{ \cos \: A} } = \frac{ {p}^{2} + 1}{2p} \times \frac{2p}{ {p}^{2} - 1 } \\ \\ \implies \sf \: \frac{1}{ \cos \: A } \times \frac{ \cos \: A}{ \sin \: A} = \frac{ {p}^{2} + 1}{ {p}^{2} - 1 } \\ \\ \implies \sf \: \frac{1}{ \sin \: A} = \frac{p ^{2} + 1}{ {p}^{2} - 1} \\ \\ \implies \sf \red{ \sin \: A = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } }$

Now,

We have :

$\: \implies \sf \sec \: A= \frac{ {p}^{2} + 1 }{2p} \\ \\ \implies \sf \frac{1}{ \cos \: A} = \frac{ {p}^{2} + 1 }{2p} \\ \\ \implies \sf \red {\cos \: A = \frac{2p}{ {p}^{2} + 1 } }$