Question

5)Ir SecA +Tand -p then find the value of 'SinA in terms of p.​

Answer 1

Correct Question:-

If sec A + tan A = p then find the value of Sin A in terms of p.

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 values from equation (1) we get,

→ p * (sec A - tan A) = 1

→ sec A - tan A = 1/p -- equation (2)

Now,

Add equations (1) & (2).

→ sec A + tan A + sec A - tan A = p + $\sf \dfrac{1}{p}$

→ 2sec A = $\sf\dfrac{ p² + 1} { p}$

→ $\sf \sec A =\dfrac {p² + 1}{2p }$

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

→ $\sf \dfrac{ p² + 1}{2p}$ + tan A = p

→ tan A = $\sf \dfrac{p - (p² + 1)}{2p}$

→ tan A = $\sf\dfrac{ 2p² - p² - 1}{2p}$

→ $\sf \large\tan A = \dfrac{p² - 1}{2p}$

Now,

Divide tan A by sec A.

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

Using sec A = 1/cos A and tan A = sin A/cos A we get,

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

Answer 2

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

Pp² - 1}