Question

If Sec
A+ tan A =p, prove that
sin A = p²-1 upon p²+1​

Answer 1

Given

secA + tanA = p

To prove

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

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

Multiply both sides with (secA - tanA)

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

⇒ sec²A - tan²A = p(secA - tan)

Now, we know that sec²A - tan²A = 1

⇒ 1 = p(secA - tanA)

⇒ secA - tanA = 1/p ......................(2)

Now, add (1) and (2)

secA + tanA + secA - tanA = p + 1/p

2secA = $\sf\frac{p^2+1}{p}$

⇒ secA =  $\sf\frac{p^2+1}{2p}$

We know that cosA = 1/secA

⇒ cosA = $\sf\frac{2p}{p^2+1}$

Put value of secA in (1) to get the value of tanA

$\sf\frac{p^2+1}{2p}$ + tanA = p

⇒ tanA = p - $\sf\frac{p^2+1}{2p}$

⇒ tanA = $\sf\frac{2p^2 - (p^2+1)}{2p}$

⇒ tanA = $\sf\frac{2p^2 - p^2 -1)}{2p}$

⇒ tanA =  $\sf\frac{p^2 - 1}{2p}$

Now, we know that

tanA = sinA/cosA

⇒ sinA = tanAcosA

Put the values of tanA and cosA

⇒ sinA =  $\sf\frac{p^2 -1}{2p}\times\frac{2p}{p^2+1}$

⇒ sinA = $\sf\frac{p^2-1}{p^2+1}$

Hence proved.

Answer 2

${\boxed{\mathtt{\red{To\:Porve}}}}$

${\mathtt{Note \: \implies \: Here \: \theta \: = \: A }}$

⇝ If sec A + tan A = p

Then sin²A = p² - 1 / p² + 1

${\boxed{\mathtt{\red{Proof}}}}$

☆ sec²A - tan²A = 1

As given that p = sec A + tan A

⇝ p² = ( sec A + tan A )²

${ \mathtt{ \red{{p}^{2} - \:1 = \: ( {\sec(a) \: + \tan(a) })^{2} \: - 1}}}$

⇝ p² - 1 = sec²A + tan²A + 2 secA . tan A - 1 .

⇝ p² - 1 = sec²A -1 + tan²A + 2 secA . tanA

$\boxed{ {sec}^{2} \: A \: - \: 1 \: = \: {tan}^{2}\: A}$

⇝ p² - 1 = tan²A + tan²A + 2 secA. tanA

⇝ p² - 1 = 2 tan²A + 2 secA. tanA

${ \boxed{ \mathtt{ \blue{ {p}^{2} - 1 \: = \: 2 \tan(a) ( \tan(a) \: + \: \sec(a) }}}}$

${ \mathtt{ \red{ {p}^{2} \: + 1 \: = \: ( { \sec(a) \: + \tan(a) )}^{2} + 1}}}$

⇝ p² + 1 = sec²A + tan²A + 2secA. tanA + 1

⇝ p² + 1 = sec² A + 1 + tan²A + 2 secA . tanA

$\boxed{1 \: + \: {tan}^{2} \: A\: =\: {sec}^{2} \: A }$

⇝ p² + 1 = sec²A + sec²A + 2 secA tan A

⇝ p² + 1 = 2 sec²A + 2tanA secA

${\boxed{ \mathtt{ \blue{ {p}^{2} + \: 1 \: = \: 2 \sec(a) ( \tan(a) \: + \sec(a) }}}}$

Dividing both values :-

p²-1/p²+1 = 2 tan A( tanA + sec A) / 2 sec A ( secA + tanA)

$\implies \: \frac{ \tan(a) }{ \sec(a) }$

⇝ tan A =$\frac{ sin \:A}{cos\:A}$

⇝ sec A = $\frac{1}{cos \:A }$

$\implies \: \frac{ \sin(a) }{ \cos(a) } \times \frac{1}{ \cos(a) }$

${ \boxed{ \mathtt{ \green { \implies \: \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } \: = \sin(a) }}}}$

${\mathfrak{\purple{Hence\:proved}}}$