Question

If sec 0 + tan 0 = p, prove that sin=p²-1/p²+1​

Answer 1

● secx + tanx = p

then

$\frac{1}{ \cos \: x} + \frac{ \sin \: x }{ \cos \: x } = p \\ \\ \frac{1 + \sin \: x }{ \cos \: x} = p \\ \\ take \: squre \: both \: side \: \\ \frac{{(1 + \sin \: x) }^{2} }{ { \cos \: x }^{2} } = {p}^{2} \\ \\ \frac{{(1 + \sin \: x)(1 - \sin \: x) } }{(1 - { \sin \: x }^{2}) } = {p}^{2} \\ \\ \frac{1 + \sin \: x }{1 - \sin \: x } = {p}^{2} \\ \\ 1 + \sin \: x = {p}^{2} - {p}^{2} \sin \: x \\ \\ \sin \: x + {p}^{2} \sin \: x = {p}^{2} - 1 \\ \\ \sin \: x( {p}^{2} + 1) = {p}^{2} - 1 \\ \\ \sin \: x = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

that's proved

● keep smiling &

always shining...............:-)

Answer 2

Answer:

Step-by-step explanation: