Question

question is given

please solve I will mark as brainlist if answer is correct​

Answer 1

Answer:

sec a + tan a = p ------->  [i]

Also,sec ^2 a - tan^2 a =1

Therefore,  (sec a + tan a)(sec a - tan a) =1        (a^2 - b^2 = (a+b)(a-b))

p(sec a - tan a) =1

sec a - tan a = 1/p------> [ii]

[i] + [ii] yields ------> 2 sec a = p + 1/p

                                            = (p^2 + 1)/p

sec a = (p^2 + 1)/2p

cos a = 2p/(p^2 + 1)         (1/sec a = cos a)

cos^2 a + sin^2 a = 1

1 - $[2p/(p^2 +1)]^{2}$  = $sin^2 a$

1 - $4p^{2}/(p^{4} +2 p^2 +1)$ = $sin^2 a$

$\frac{[(p^4 + 2p^2 + 1)] -4p^2}{(p^2 + 1)^2}$ = $sin^2 a$

$\frac{(p^4 - 2p^2 + 1)}{(p^2 + 1)^2}$ = $sin^2 a$          (a^2 - 2ab + b^2 = (a-b)^2)

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

Thus, sin a = $\frac{(p^2 - 1)}{(p^2 + 1)}$

Hope it helps!!!